 So permutations and combinations, permutations and combinations. This chapter is basically not completely new to most of you because even in class 10th or as early as class 9th, we were basically taught permutations and combinations. Of course, in permutation and combination in class 9th and 10th, when you're junior classes, you basically count by literally making a set out of it, right. But now in 11th and 12th, the scope will be much broader where we'll be using certain principles to count. Okay, we'll be talking about all those in our session today. And this topic is especially useful for the people who are preparing for KVPY. Okay, some news was there that KVPY has been called off by the High Court orders. We are waiting to, you know, get more information on this news as and when we get it will keep all of you updated with respect to the news. And another important thing about this chapter is two other chapters are directly linked to this particular concept. One is your probability and probability is there for you in class 11th and 12th both. Okay, and one more chapter which is required for your competitive exam that is your binomial theorem, even though it is not there for your CVC curriculum. And that chapter also is very closely linked to permutation combination. Not everything would be required, but yes, the basic idea of permutation combinations will be used there as well. Okay. So here, first of all, let's understand the literal meaning of these two words. Right, what is the meaning of permutation? Have you started this chapter in school? Anybody who has started this chapter in school? Can I hear from you? What is the meaning of permutation? Permutations, permutations means? Arrangements, right. Arrangements. What is the meaning of arrangements? Does it mean ordering? Does it mean selection or does it mean both? Arrangement is an English word, right? So what is the meaning of that? Only selection. Okay. Only ordering. Yes, exactly. Arrangement is basically a combined activity of selection and ordering both, right? So let us say there are five people, or let's say there are 14 people. Nowadays, ICCT20 World Cup is going on. Let's say out of 14 people, sir, don't talk about it, India lost to Pakistan, right? How many of you wasted that three and a half hours watching that match? I'm not happy that India lost. But basically, he should not be spending so much time watching a match, okay? Anyways, so let's say there is a 14 member squad and you want to select 11 people and assign them some order of batting, right? That is what we call arranging 11 people taken from 14 people, okay? So in arrangements, we do both the activities. Ayyo, Seth. We do selection as well as ordering, selection and ordering. So don't worry about spending and losing that much, okay? I think it's the first time Pakistan won against India in a T20 World Cup, right? Okay. But on the other hand, what is a combination? Combination is just selection, okay? It is just choosing. No ordering is done in case of a combination. So combination is just selection, okay? So how many selections or how many choices can be made? For example, if I ask you to just pick up a team of 11 from 14. No abatting order is to be assigned to them. Then that is a combination, okay? Now many people ask me this question, sir. In permutations, you have selection and ordering both and in combination, there's only selection. So why do we study permutation first? Why is the chapter name is even permutation and combination? They should call it as combination and permutation, right? Because permutation is a two-phase event. We have selection and the other phase is ordering, right? But in combinations, we only do selections, right? So why don't we study just combinations and then permutations? See, it is because permutation is more close to our fundamental principle of counting. That is why permutation is studied before combination, okay? So we'll be talking about fundamental principle of counting also in today's session. But before anything, I would like to talk about factorials because factorial is something which is going to be a tool in your hands while you're solving or doing a lot of calculations, okay? So it's good that we have some bit of exposure to factorials. I know you have all done factorials with me during the bridge course. What is the difference between these two? A lot of difference is there, Sethu, see? Let me give you a simple example. Let's say I say abc, okay? D, I ask you to select three alphabets from abcd and arrange them, okay? So what could be your possibilities? So if I talk about arrangements, your possibilities for arrangements would be the following. You have abc, okay? You will have, see, now you are selecting, okay? I think I have taken a lot of alphabets. Let me take three only, okay? And let me say I want to arrange arrangements of two alphabets or two letters from abc, two letters from this abc, okay? So you can have ab, you can have ba, you can have ac, you can have ca, you can have bc, you can have cb, okay? So there are six arrangements possible of two letters from abc. But if I say combination of two letters from abc, so if I have combinations of two letters from abc, two letters from abc, then you can have only the following combinations, ab, ac, bc, okay? You don't have to order the choices that you have made. You just have to select them and the moment you select them, the combination is completed, okay? So these are your combinations. So there are only three combinations here, but there are six arrangements of two letters taken from abc. Because here ab and ba, since you have changed the position of a and b, that is taken as a different ordering. Combination wise, it is the same combination, but the ordering of a and b here has mattered. That's why I told you, let's say if you are selecting a team of 11 people, okay? The selection is combination. But now if you decide who's going to bat at number one, at number two, maybe who's going to open, who's going to come one down, two down, three down, that is basically your ordering part of it and that comes under arrangement. Subset, I can't say that, but yes, it's an activity which may proceed, okay? You'll come to know when I talk about both of these topics in detail. So normally it's not that you always select, but yes, there is a choice involved if you want to, okay? We will talk about it. Don't worry about the integrities of these two topics. These topics will come little later on. Before that, we have a lot of prerequisites to talk about. And the first of them being factorials, okay? So what is the meaning of a factorial? Factorial is nothing but it is a continued product starting from the number and till you reach, till you reach a one, okay? Factorial is always found for the whole number, including zero, right? It is not defined for negative integers. It is not defined for fractions, okay? In older books, you can see that this representation was used for factorial, especially if you look at higher-algebra older books, like Burnard Child, Hall and Knight, they use this kind of representation to represent factorial. Some examples of factorial I'll give you over here. So let's say if I talk about five factorial. So we know it's a continued product from five all the way till you reach one, okay? Five factorial is nothing but 120, okay? Four factorial, four factorial is four into three into two into one. Now, what is important here to note is that four factorial is obtained by dividing five factorial by five. That means the higher factorial, one more than this, if you divide by the same number, you get the next factorial. This is a very important property. I will be talking about that also. Similarly, three factorial will be three into two into one, which is actually nothing but four factorial by four. Similarly, two factorial is two into one. That's nothing but three factorial by three, okay? And one factorial is nothing but one, which is two factorial by two. And now zero factorial basically, you know, leads to some kind of inconsistency in the definition. The definition says you need to start from the number and go all the way till down to one. So zero factorial comes as a mathematical inconsistency over here. Now, in order to resolve this inconsistency, it was basically decided to follow this pattern that I was using up till now. So as per this pattern, the answer should be one factorial by one, right? And that also comes out to be a one. So note this down, zero factorial is also one just like one factorial is. But remember, this is just to fill up that mathematical inconsistency. I can't go to minus one factorial because if I try to go to that, basically I will be doing zero factorial by zero and dividing by zero itself is not defined. Now the moment minus one factorial becomes undefined, the further down I cannot go. I cannot go to minus two factorial, minus three factorial. And that's why I was trying to say that factorial only works for whole numbers. Is this fine? Any questions? Now, normally people ask me, do we have to remember these factorial values? Up till seven factorial, you should remember. Five factorial I've already written. Let me write down the six factorial also. Six factorial is 720. And seven factorial is 5040. Beyond that, you don't need to remember. So till seven factorial, you should all remember. Any questions? Now, what is the physical significance of a factorial? Let's talk about that and we'll also look into certain properties of factorial. If you have any questions with respect to what is a factorial definition, et cetera, please highlight. All good. Any questions? Anybody? Okay. So what is the meaning of factorial? The physical significance of a factorial is, and I will be talking about this in our later part of our discussion, factorial actually represents the number of ways to order and distinct objects among themselves or taken all at a time. So if you have, let's say, five objects, all of them are distinct. This word, this distinct word is very important. If they are not distinct, you cannot use n factorial for that. So let us say there are n distinct objects given to you and I asked them to order them among themselves. That means place them in all different orders that you can. When I say order, basically you're trying to arrange them in a linear order in a row wise fashion. So if you try to assign a linear ordering to n distinct objects, the number of ways you can do that is actually n factorial. Let me take an example. Let's say A, B, C itself. In how many ways can you order A, B, C? Tick in all at a time. So you'll say, sir, A, B, C, A, C, B. Then you'll have B, A, C. That's beyond my ability. Yeah, I know it's beyond your ability. Okay, B, A, C. Then you'll have B, C, A. Then you'll have C, A, B. And then you'll have C, B, A. So the number of ways you can order A, B, C among themselves or taken all at a time is basically nothing but three factorial. Okay. We will be talking about this when we talk about permutations. Okay. So this is one important physical significance of factorial. Now a few properties of factorial that you should be keeping in mind. The first and the most foremost property that is going to be handy in solving most of the question is a factorial of a number is basically linked to a factorial lesser than that number by this relation. For example, five factorial was nothing but five into four factorial, isn't it? So this particular relation is something where you are linking a higher number of factorial to a lower number of factorial. Okay. Or in other words, you're writing a higher number of factorial in terms of a lower number of factorial. Now this can go on and on. For example, I could still break n minus one factorial as n minus one times n minus two factorial. I can further break it down as n minus one n minus two times n minus three factorial. Okay. And this can go on and on. Now many people ask me, sir, you know, how do I know till where we have to stop? It all depends on the question. For example, let us say somebody has asked you. I'll just take an example question for this property. Let's say somebody has asked you to evaluate 22 factorial by 20 factorial. Right. Now, of course, you will not sit and multiply all numbers starting from 22 all the way till one and 20 all the way till one, because you know that even if you multiply and you divide it, a lot of terms are going to get canceled off. Isn't it? And of course it will look very awkward to write all those numbers and start cancelling it. Right. Looks very condition. So what we can do here is 22 being a higher number, 22 factorial can be written in terms of 20 factorial like this 22 into 21 into 20 factorial. So, you know, you have to stop till 20 factorial. So as the case may be, you can stop till that factorial and you can just, you know, conclude your result. By the way, how much is this? 22 into 21. 42, 42. I think it's 6, a 462. Is it fine? Any questions, any concerns with respect to this property? So you can evaluate this pretty quickly. Okay. Next property, which I would like to, you know, mention here. In fact, it's a mention, not a property. Please understand this. Such properties do not work on factorial. Don't try to do that. X plus Y factorial doesn't mean X factorial plus Y factorial. Same goes with XY factorial. It doesn't mean X factorial, Y factorial. So don't try to extrapolate these formulas from your end. Let's say X by Y being a whole number. Let's say X by Y is a whole number and somebody says, you know, evaluate this. It doesn't mean you do X factorial by Y factorial. Please. I know I should not even mention these things, but there were people in past who misuse these properties like this. Kindly do not do that. Okay. Next thing is if X factorial is equal to Y factorial. It either implies X is equal to Y. Or X is zero. Y is one. Or X is one. Y is zero. That's the only two possibilities. Okay. So if X factorial and Y factorial, I mean, I'm just giving you some scenario that if they're equal, either X and Y are equal. Okay. Either an XY are, you know, X and Y are, let me make it here. Yeah. Either X and Y are equal whole numbers or they are related to each other, like X could be zero. Y could be one or X could be one. Y could be zero. Okay. Is this fine? Any question with respect to what is a factorial? We'll talk more about factorial in our upcoming concepts, which is called the Lujans formula. We'll talk about Lujans formula in some time, but before that I want to take some basic questions on factorial. Hello, good evening. Those who joined in late. We have started a new topic today, permutations and combinations. Right now, if you see, we are on the very first slide and we have started with the discussion on factorials. Okay. Anything that you would like to copy here, ask here, please do so because now I'm going to move to the next slide. All happy online. So let's move on to the next slide and I will take a few questions with you all. Let's take a few questions. Where is my in class sessions in class? I would like to start with this question. This question says prove that. The expressions on the left-hand side simplify to 2n plus one factorial by n factorial. You may start, you may do the proof from any direction. You can do from right to left or left to right. That all depends on you. So please do this and do let me know when you're done. Just like you're done on the chat box. Anybody. Oh, yeah, take, take your time. No issues. Nikhil is done. Wonderful. Nikhil. Done. Shardhuli also. Very good. What about others? Dear all, everybody, please participate. Adya, Arif, Abhijay, Anushka, Arundhati, Harshita, I don't want to name everybody, but you know that you need to participate. Okay. I can only see a few, a handful of you participating. Yeah. Okay. So let's start this proof. Maybe I'll start with this expression. Two n plus one, the whole factorial. Okay. Now, if you see two n plus one whole factorial is basically nothing, but it's a continued product. Okay. It's a continued product till two n plus one. Okay. Now, from every event term that you have over here, do one thing for separate the event terms and the odd terms from each other. Okay. So first I will write all the event terms, four, six, eight, da, da, da, da, da, da, da, da, till two n and all the odd terms. And let me use white color font to write them. Da, da, da, da, da, da, till two n plus one. Okay. Now, from each of these terms that you see in the yellow, which is your event terms, start pulling out a two, two, two, two, two each from each of the terms. Right. So if you pull out a two from two, you get a one. If you pull out a two from four, you get a two. If you pull out a two from three, you get a, sorry, pull out a two from six, you get a three, and so on and so forth till you pull out a two from two and you get an n. Now, when you pull out the twos from each of the terms, you know that there are n number of terms here. So ultimately you'll get two to the power n, leaving behind one, two, three, four till n. That is actually n factorial. We'll write that down a little later on. Not a worry. So this white terms, I will write it. I'll write them as, as it is. Okay. Okay. Now see here, automatically these terms will start, you know, shaping up as the required expression that we need. So this is nothing but n factorial. Okay. And these terms are already present in your expression. So I will not worry too much about these because they are definitely there in your expression. Okay. Now from here, the final step is bring down that n factorial to the left side. And that's how your expression becomes, that's how your left-hand side and right-hand side become equal to each other. Hence proved. Is it fine? So we just use the basic definition of factorial and just, you know, manipulated with the terms and got my result. Fine. Easy. Any questions, any concerns? Could you take another one now? Okay. So let's take another one. Feel free to stop me if you're copying anything. Okay. So when you don't say, Karthik is saying, how do you know how to separate odd and even? Why? Because I saw the expression that only odd terms were seen there, Karthik. So when you're solving a question, have an eye on what you want to achieve. Okay. From there, I got the idea. Is it fine? Karthik, all good? Okay. Let's take another one. Let's take another one. Okay. Yeah. Maybe we'll take this question also. Solve for X. Solve for X. So it is given that X factorial upon two factorial X minus two factorial is to X factorial upon four factorial X minus four factorial is two is to one. Solve for X. Solve for X. I would expect everybody to reply to this on the chat box. Don't worry about answer being right or wrong. Participation is more important. Yes. Yes. It is to symbol. Why? There was any doubt related to that? So read it like this, Karthik. This is to this is two is to one something like that. And one more thing I would like to ask you here is that till what time does your school operate in the hybrid mode? Till what time was the last period? One o'clock. Okay. So you had enough time to reach home. No, I hope you're not. You're answering to the timing, right? Okay. One PM was the last. Okay. Okay. So, okay. We'll check whether there is an answer to that or not. Vashna. Okay. Very good. So Siddharth, Karthik and Vashna. Yeah. I saw your answer. Both of them. I saw your answer. Anybody else? Sharduli also I got your answer. She has a different opinion. Anybody else? Nikhil, Nikhil, Nikhil, Nikhil, Nikhil. Where was your answer? Okay, Nikhil. Your answer is also okay. Okay. Anybody else? Okay, fine. Let's let's do this. See, first of all, even before you try to solve the problem, please understand factorial works only for whole numbers. So some of you who are giving the answer like half or zero. That answer cannot be your answer because if you had 0 as then your denominator here would be minus 2 factorial which is not defined okay or minus 4 as your answer which is not defined or negative 3 by 2 factorial not defined okay so the moment you get a value of x see they might appear in your answer I'm not saying they will not appear while you're solving it but it is up to you to rule them out you will have to discard those answers because those will not make those factorial terms you know actually defined so such terms would be neglected that would be ignored that will be rejected okay so okay okay let's discuss it out yeah these two means you're dividing this term by this term so let's do that so you're dividing this term it will be x factorial 4 factorial x minus 4 factorial now as I told you factorial basically allows you to write a higher value factorial in terms of a lower value factorial okay so when you do that you can actually write 4 factorial in terms of 2 factorial but many of you would say sir why to do that this is 24 this is 2 we already know it so make it 12 good save your time no issues okay and here you have x minus 4 factorial and x minus 2 factorial now remember x minus 2 is a bigger number as compared to x minus 4 so you can write x minus 2 factorial as x minus 2 x minus 3 into x minus 4 factorial right just score these two off okay and now you have x minus 2 x minus 3 I am taking it to the right hand side and I'm bringing this 2 down under 12 which is 6 now many of you would have got a quadratic expression over here okay but you don't have to get into quadratic also for solving this now listen to this very very useful piece of advice that I'm going to give you here you know that these are two consecutive numbers right x minus 2 and x minus 3 are consecutive okay this is the higher this is the smaller of the two so right and both of them have to be both of them have to be natural numbers remember that remember x itself has to be a natural number or whole numbers to be more precise for x factorial to be existing isn't it so try to break 6 as product of two consecutive whole numbers and the only possibility to do that is 3 into 2 which means you can do a direct comparison you can directly say x minus 2 is a 3 or you would say x minus 3 is a 2 it doesn't make any difference either of the case you will get x value as a 5 only so this is the only possibility this is the only answer any other answer that you would write would lead to it being wrong no worries sharduli it's okay karthik clear everybody is clear okay let's take another one anything that you would like to ask or no please do ask this this approach which I've told you in the last step that is very important and very handy to solve few of the questions little later on I will take that up in later part of the topic okay so next question that you are going to take up is here I'm going to change a question slightly I'm going to first ask you to do the sum till 1 to let's say 50 1 to 50 n is 50 here I will come to n also don't worry but as of now I would like you to sum up r into r factorial from r equal to 1 till 50 r equal to 1 till 50 question is clear so question is I wanted to sum 1 into 1 factorial 2 into 2 factorial 3 into 3 factorial all the way let's say till 50 terms which is 50 into 50 factorial so what is the answer here that is what I need to figure out of course your answer should also our answer will be in terms of factorial only don't try to literally add them because these numbers are too big okay excellent Nikhil very good awesome great your answer is absolutely right anybody else now I'm sure most of you would be struggling to you know start actually in this right because you know you you have definitely done your you know summation chapter the series topic that we did in the last class but even you know with that knowledge how do you do this because the factorials are coming into picture now not to worry there is one special method which I discussed in the last class which was called the vn method right I'll be using my vn method in this scenario as well let's see how 0 are a why such a big number you're calling a zero all of them are positive why they should add up to give you a zero okay never mind we'll check it out see so first of all write down the rth term over here the rth term over here is r into r factorial which is already given to you in the expression itself now write down this r as r plus 1 minus 1 okay and let's expand this by you know clubbing r plus 1 together and this one is separate so when you multiply it gives you r plus 1 times r factorial minus r factorial right now what is r plus 1 into r factorial what is this r plus 1 into r factorial r plus 1 factorial correct now here is the thing which I was basically talking in our vn method of summation last class that basically I've expressed r rth term as difference of two systems which are coming from the same function where one has an r another has an r plus 1 or r minus 1 so in this case something similar to that has happened so what is the advantage of writing r into r factorial like this now you yourself can see here when you start putting your r value as a 1 let's say when you put r as 1 you get this as 2 factorial minus 1 factorial when you put r as 2 you get this 3 factorial minus 2 factorial when you put r as 3 you get this as 4 factorial minus 3 factorial now let's say I want to go till the 58th term so let's let's write 49 term over here which is going to be 50 factorial minus minus 49 factorial why did we write r as r plus 1 I never wrote r as r plus 1 I wrote r as r plus 1 minus 1 that's r right Pratej isn't r plus 1 minus 1 and r correct now if you're asking what was the reason for writing it like this is because my purpose was to break it up as difference of two factorials both coming from the same functionality right okay yeah so the next the last term that would be t50 would be 51 factorial 51 factorial minus 50 factorial now let's add them that's what we are basically you know looking for we are looking for the sum of r into r factorial from 1 to 50 as you can see there would be a cancellation of these alternative terms okay ultimately leading to I mean terms in between dot dot dot they will also get cancelled off so ultimately leaving you with 51 factorial 51 factorial minus 1 factorial right and there you go the answer for this particular series is going to be 51 factorial minus 1 now can we generalize this can we generalize this act and try to answer the initial question which I had given where there was an n here so in that case what would be your answer please write down everybody so in general if you try to sum up this from r equal to 1 to n what should be the answer coming up absolutely n plus 1 factorial minus 1 is it fine any questions any questions any concerns with respect to the solution okay now the next important concept which is actually associated with factorial is the lujan's formula let's talk about that anything that you would like to copy ask in the slide please do so great okay let's move on to the next one so the next thing that we are going to talk about is is now you can see on your screen I have written a spelling legendary but it is not pronounced like this okay so it is pronounced as lujan lujan okay lujan's formula so what is this lujan formula lujan formula is basically a formula given to us by a french mathematician okay of course his name was lujan where he basically helped us or where basically he discussed how to find out the exponent of a prime number exponent of a prime number in factorial of a number okay so basically he basically gave an idea or formula for where we can actually figure out that if you prime factorize a factorial of a number then what would be the power exponent exponent is power so what would be the power of any of the prime numbers in the prime factorization I'm sure most of you would have prime factorized numbers in your junior classes right we normally call them as writing it as a canonical form okay so let's take an example maybe let's say four factorial four factorial we know is 24 okay if you prime factorize 24 what do you get you'll say I get 2 to the power 3 into 3 to the power 1 correct so lujan formula basically helped you to find out what would be the power on any prime number which basically occur in the prime factorization of the factorial of that number okay and he basically gave a symbol for this he called it as e p n factorial just to make it a notation e means exponent p is the prime number and n factorial of course the number from where you are trying to find the exponent so if I say what is the exponent of 2 in 4 factorial what will you say what will you say what is 3 correct what is the exponent of 3 in 4 factorial what will you say 1 okay now of course this number was small enough and you could factorize it and figure it out but what if the number is big enough let's say I gave you 100 factorial and ask you hey what is the exponent of 2 or maybe 3 or maybe 5 in 100 factorial how would you figure that out for that lujan gave a formula to us our assumption no it's an example no don't say don't call it as an assumption approach we took an example to this is an example which I'm taking all right okay so now the formula basically evolves like this I mean I will be writing down the formula and then we will justifying why he or how he got that formula actually so lujan said that the exponent of any prime number p in factorial of a number is given by gif of n by p plus gif of n by p square plus gif of n by p cube till you start getting zero okay till you start getting a zero so there would be a case where your p cube or sorry p to the power some power s will exceed n in that case you'll start getting zero zero isn't it okay so here this square bracket that I have written basically represents the greatest integer function gif greatest integer function okay now how possibly would he have got this formula let's try to understand from this example itself okay so in four factorial what is the exponent of two let's try to figure it out I know you have already you know written the answer three but let's try to figure it out so four factorial is what four factorial is one into two into three into four okay so what do you do to figure out how many twos are there in this product what is the common sense you know say in order to know how many twos are there in this product what are you going to do the first thing that I would do is I will pull out the two from every even number correct because I know every even number will at least contribute one two isn't it so there is an even number two here there is an even number four here I will take out one one two from both of them correct so this will contribute a two this will contribute a two right so I already have I already let me just keep a tag of the number of twos so I'll only get one two from here and one two from here correct right but remember there are there is a number over here which has actually got two twos isn't it and that number here is four so that number will contribute one more two right in short can I say all multiples of two is contributing one one two at least all multiples of four will contribute one more set of twos similarly all multiple of eights will contribute further one more set of twos and so on and so forth correct isn't it so one one two you already got from here and this guy four will give you one more two isn't it so now this two that you have got it's actually the number of factors of twos that you have in this particular continued product and that is only two in number if you write gif that doesn't change the value it is still a two but how many fours you have from one to four you will say it's the only one which is actually four by four and four by four is two factorial and that is what basically your lujan formula also gave you right and that's how you get your answer as a three let me take one more example okay from that example further things will be clarified okay let us say my question is find the exponent of find the exponent of two in let's say uh 12 factorial okay so find the exponent of two in 12 factorial so how will I find this out okay now I'm solving this problem from common sense point of view I'm not using the formula so let's try to see whether the formula is coming from some logic or not so if you start writing 12 it's not many numbers I'll write it down no worries okay so first I'm going to do I'm going to pick out 111112 from every multiple of twos how many multiples of twos will be there from one to 12 how many multiples of two are there five count one two three four five six uh six isn't this six comes actually when you do 12 by two isn't it okay now gif is put just to ensure that we are not taking this you know if let's say fraction comes fraction should not be taken into account okay so six twos I already got okay so think as if I make a very you know interesting scenario out of it let's say you are a tax collector okay your tax is the number two so you're collecting two from the people who can provide you two okay so first what did you do you said you asked all the even numbers hey you all have a two right give me two okay so you're like you're like collecting the tax and the taxes that twos okay so these even numbers they will say okay take my two take my two all the six numbers give you the two two right but there are some rich people also who have more than one twos like four eight twelve correct these guys will have two twos right so they can further give you more twos that's what happens in our country correct rich people have to pay more tax isn't it poor people don't pay more tax correct right that is a socialized what do you call this uh socialistic pattern of governance right so these numbers four eight twelve they are rich people in terms of two so they can give one more set of twos so they have already lost a two but they still have more two with them so more money is there in their pockets so government will try to take that out also right so they will further lose one more set of twos and how many set of twos will come from here as many number of factors of four are present so how many factors of four are present for one to twelve you can literally count it actually four eight twelve three which is equivalent to saying 12 by two square okay is it fine any questions but there are some people who are having more twos this guy eight this guy has got three twos so even after losing one hair and after losing one hair it is still having more money with him or more twos with him so do you think government is going to leave him no more you know they have to give more taxes so you have to find how many factors of eights are there and how those many factors of eight will contribute one more set of twos so you just keep on adding it okay now any more you know factor will not be there because you know they have all lost all the twos that you can give so literally counting a number of twos will be six plus three plus one that means 10 twos can be picked out so if at all you decide to prime factor is 12 factorial then two will have a power of 10 of course there will be some power of three there will be some power of five I'm not going getting into details of it my question was only to get what is the exponent of the prime number two in the prime factorization of 12 factorial and that answer is 10 okay now here literally you can count and verify so this has one two this has got two twos this has also got one two this has got three twos okay this has also got one two this has got two twos so literally count one three four seven ten ten is the answer okay so this whole activity the french mathematician loujon converted like a formula like this way and he made our life very very easy so you don't have to literally see find out all those things is the you know justification from why he gave this formula clear to everybody some people are joining late please understand the seriousness of these classes joining late means you will not get the context of what is going on right people who have joined now two of you have joined now I don't know who all you will not understand what's going on okay see saithu you need to collect the twos from all the terms which can give you two and you have to add how many twos you have collected that is the exponent of two in a number right correct have you seen that old two rupee note nowadays many people don't use it now there are coins no okay so when we were small no my my mom used to give me two rupees okay for you know getting some kind of snacks in the school okay but my dad used to give me five rupees okay because of course big heart my dad so if let's say you know there are 12 students okay and they are carrying two rupees with them okay so this guy only has one two rupee think like that okay and this guy has got two two rupees this guy has got three two rupees this guy has got one two rupee one three rupee like that okay anyways so let's say you are a person who is coming to take taxes from them and you you are collecting only two rupees from them how many two rupees can this guy provide you he'll say only one two rupee I have take it this guy can provide two two rupees to you this guy can also provide one two rupee to you only because he has got one two rupee and one three thing like that okay so like that you are collecting all the twos that you can collect from these numbers and trying to see how many twos you have collected that is what we are trying to find out in this topic what is this formula for Tejasani for a bigger number this formula will be used these are all smaller versions of the same formula right let's take a question let's take a question then you will understand how you can collect the twos from a bigger number okay let's say I ask you what is the exponent of two from hundred factorial okay find exponent of two from hundred factorial of course you can't sit and multiply till hundred so for that we will use the Lujans formula as per Lujans formula the number of twos that you will get would be the value that you are going to get from this operation so Lujan formula is a way to figure out those value that exponent isn't it okay now tell me where to stop because I don't want to unnecessarily write terms which will give you zero should I stop here or should I go further 2 to the power 7 is 128 so if I divide 100 by 128 I'll get a fraction and gif of a fraction will be zero isn't it so now count gif of 2 by 50 sorry 100 by 2 is 50 this is going to be 25 now there's an easier way to calculate it you don't have to always sit and find this value just keep dividing the previous answer by a 2 and take the gif so 25 by 2 is 12.5 take a gif it'll be 12 12 by 2 6 6 by 2 3 3 by 2 gif will be 1 okay so if you add it you'll end up getting I think 75 87 plus 10 97 okay so just any had you factorized hundred factorial as 2 to the power something 3 to the power something 5 to the power something then that 2 there will have a 97 on the power understood how does this formula work any questions anybody okay so if you're clear with this let me ask you a question find exponent of 5 in 100 factorial find exponent of 5 in 100 factorial very good Nikhil very good Vashna excellent very good so the answer will be simple 100 by 5 100 by 5 square now beyond that I need not go because 100 by 5 cube will lead to a zero okay so all the other terms will be zero so it's as good as saying 20 plus 4 which is 24 okay so this this term which I had left here actually this will be 24 okay now use this result the two result and tell me tell me the answer to this question tell me the answer to this question how many zeros are there at the end of hundred factorial use the two results which are just now written over here excellent Nikhil I didn't catch that could you try again correct very good 121 oh my god why correct Seethu correct Shalini okay the answer to this question is 24 again why now see very simple in order to make a zero at the end of a number that it means how many tens you have multiplied it with correct so let's say the count of the number of zeros at the end of the number is basically how many tens are there in the you know that number that means how many tens are required or how many tens are getting formed in that number isn't it now in order to make a 10 you need two prime numbers one is two another is five correct so in order to make one 10 you need one two and you need one five nothing as if you're talking you're talking in terms of chemistry limiting reagent I'm sure you have done a stoichiometry right concept of limiting reagent what is a limiting reagent so when there is when there are multiple reactants which are giving you a product right and of course you have been given some quantities of those reactants then the one which is then the one which is in a smaller quantity that is the limiting reagent because that fellow decides how much product will get formed isn't it so now think as if one mole of two and one mole of five react to give you one mole of ten okay and in hundred factorial there are 97 moles of two and only 24 moles of five then how many moles of ten will be created 24 moles of ten because this is the limiting reagent correct so I'm talking in terms of chemistry I hope that makes sense to you so out of two and five five is in a lesser quantity and it will always be in a lesser quantity because the higher number will occur lesser number of times in the factorial of a number okay so in case of hundred factorial the number of zeros at the end will be decided by how many files are there in the prime factorization of hundred factorial and there are only 24 so since there are only 24 this is your limiting reagent limiting reagent I'm writing it down in terms of you know chemistry language since this is a limiting reagent there will be only 24 24 10s created okay so if there are only 24 10s created then hundred factorial will end in 24 zeros zeros you can write it as you are also is it fine any questions okay so just to test you on this concept further I will give you one more question can I go to the next slide find exponent of of 12 in 50 factorial okay Vaishnav okay Setu okay Shalini three answers so far four answers so far very good Shardili Arav anybody else who would like to contribute okay now I'm getting three types of answers in fact I'm getting uh mostly two types of answers okay let's discuss them oh yeah why not Vaishnav if you want to change your answer please go ahead no issues do it do it do it okay now some of you are giving me the answer for this as four maybe because you're doing this okay that's why you're writing your answer as a four this is can you tell me why it is wrong it's because you have not read the name of the topic properly let me go back read the name of the topic once again loudly exponent of a prime number in n factorial my dear students is 12 a prime number for you huh that's what fair attention see these are small things where you will be caught off guard okay 12 is not a prime number this formula is not going to work now why is this formula not going to work why did I just cross it out is because see a prime number cannot be made from several factors other than the number itself and one but a non prime number let's say I want to make a 12 I want to count how many 12s can be made 12 is not only made by numbers which are multiples of 12 or let's say multiples of 12 square or whatever they can also be made by multiples of let's say three and four correct are you getting my point because of those issues coming up you cannot use this notation to get your exponent of 12 so first of all what is the meaning of exponent of 12 and 50 factorial that means how many 12s you can get from 50 factorial right if I have to extract all 12 12 12 12 from 50 factorial of course leaving behind an integer how many such 12s I can pull out that is what the question is asking so here the concept is same as what we did for no finding how many 10s are there in 100 factorial so 12 as a number is what 12 as a number is 2 into 2 sorry 2 to the power 2 into 3 to the power 1 right so see what prime numbers are involved in making a 12 only 2 and 3 are involved in making a 12 isn't it and that too 2 moles of 2 again I'm using chemistry language 2 moles of 2 and 1 mole of 3 is required to make 1 mole of 12 okay now see first figure out how many 2s are present in 50 factorial let's answer this question you will say sir simple now I can use my Lujan's formula so it's gif of 50 by 2 gif of 50 by 2 square gif of 50 by 2 cube gif of 50 by 2 to the power 4 gif of 50 by 2 to the power 5 I think beyond this I need not go because 2 to the power 6 is going to be 64 and that's going to give me a 0 0 so this is 25 this is 12 this is 6 this is 3 this is 1 how much 37 37 was 10 47 so they are 47 2s present in 50 factorial how many 3s are present in 50 factorial let's figure that out as well so that's going to be gif of 50 by 3 gif of 50 by 3 square gif of 50 by 3 cube I think beyond that I need not go because 3 cube is going to be 90 sorry 3 to the power 4 is going to be 91 sorry 81 and that's going to give me a 0 so this is as good as a 16 51 correct so that's going to be 22 in number okay now again ask yourselves 1 mole of 3 and 2 moles of 2 make 1 mole of 12 you have 22 moles of 3 you have 47 moles of 2 which of the 2 is the limiting reagent right which of the 2 is the right this is the limiting reagent right so 22 3s will actually require just 44 2s to make a make 22 12s right and 2 is in abundance we have 47 2s present there are only 24 of them required okay so 22 3s require 44 2s to generate 22 12s so the exponent of 12 even though you know we are writing this expression so the exponent of 12 in 50 factorial will only be 22 because only 22 12s can be generated okay remaining 3s remaining 3 2s will be wasted not not be utilized in making a 12 is it 5 so the answer to this question is 22 the first one to get this right and let me name the student the first one to get this right wash I think Sharduli yeah Sharduli and Aarav Sinha excellent very good won't this be an exponent of 6 6 also will be you know 22 only because to form a 6 you need a 13 and a 12 okay so 22 22 6 can only only be formed so if somebody asked you what is the exponent of 6 in 50 factorial that will also be 22 okay clear yeah it's okay is it fine great now with this I'm going to begin with our main part of our chapter the basic part of our chapter and that is called the fundamental principle of counting fundamental principle of counting now these principles are called fundamental because they are very fundamental to our logical way of counting right so so fundamental that you can't explain it also why we are doing it like this okay so it's like you know fundamental forces like gravitation is one of the fundamental forces right we know there is a gravitational attraction between two celestial bodies or any two bodies for that matter but why do why do they attract each other why there's a gravitational pull between two bodies very difficult to you know explain the scientists are still working on it say to have some question let's resolve that first and 11 for 12 because there are 22 twos so there are 11 force see to make 112 you need two twos right and you need one three correct no and you have been given 47 twos and only 22 threes so those 22 threes will will react with I'm using the chemical language you react with 44 twos to generate 22 twos what is the doubt there the that four will that four that will be left off you're talking about the three twos that will be left off right these three twos do not have any threes left to make a 12 ah now you're interested okay now let's talk about fundamental principle of counting so primarily we're going to talk about two fundamental principle of counting which is fundamental principle of addition and fundamental principle of multiplication however there is one more interesting fundamental principle which is called the principle of inclusion exclusion which will be going we'll be talking about later in our chapter much later when we talk about distribution of distinct objects in different groups and also when we talk about derangement concepts now what is this fundamental principle of addition what is this addition fundamental principle now see I'll give you a simple scenario very simple scenario let us say there is a you can say movie theater okay and you are inside the movie theater okay now after we the movie is over there are multiple exit gates right you would have all the experience it some people are living from front some people are living from back some people are living from sides multiple ways they leave okay so let's say there are three exit doors from here and there are four exit doors from here okay tell me if there are three doors from here and there are four doors from here how many ways can you exit the cinema theater how many ways can you number of ways to leave the a number of ways to exit the cinema theater what is the number of ways seven seven is nothing but three plus four isn't it you added three and four right you got seven and they answer now do you have any answer why did you add them why not three to the power four why not four to the power three why not three into four why did you add them now I know these type of questions are difficult to answer because they are so fundamental isn't it some fundamental things are very difficult to explain okay now basically you say sir I could do this task of you know getting out of the cinema theater in the following ways I can go out of this door or I could go out of this door or this door or this door or this door or this door or the moment you're saying or logically it means you have to add okay so it's a logical connection by which you are adding it right there's no hardcore derived formula that okay this this means it has to add okay so intuitively or logically speaking you are adding it isn't it okay so this thing has been put as a principle and the principle says that if there is a task okay if there is a task which you can accomplish by doing any one of these let's say no cases let's say case one case two case three etc I'm just putting some dot dot dot okay so let's say case n so if you're basically trying to do a task by doing any one of these you know cases independently then the total number of ways of doing this task will be the sum of the number of ways of accomplishing these cases right so let me write this in the English language so that later on when you're referring to your notes you're clear of what basically I'm trying to say here so let's say if your task if your task what happened to my marker yeah if the task can be accomplished accomplished by independently performing the subtas under the following cases one two three and so on okay then the total number of ways then the total number of ways number of total number of ways of completing task completing the given task is equal to sum of the number of ways case one case two case three tasks okay so here what was your task your task was to get out of the cinema theater right so there are several cases you can get out of the cinema theater if you step out from door number one that's case one or door number two case two or door number three case three or door number four case four like that so you can also say that you can take any one of the doors to your left okay so there are three such ways to do that task or you could take any of the doors on the right of you there are four doors present there so let's say I briefly divided into just two cases case one and case two so case one is taking a door to your left case two is taking a door to your right so taking a door to your left you have only three ways taking a door to your right there are four ways to do it so three plus four that is the sum of the ways that in which in which you can do case one and case two if you add them that will give you the total number of ways of doing that you know task of getting out of the cinema theater okay now these all things look very simple but when it comes to application in questions many people like confuses so do I add them do I multiply them all these kind of questions start coming up looks very trivial very simple but when it comes to its application people do lot do lot of mistakes is it fine okay now let me take another question here let us say let us say in your classroom there are five boys and let's say six girls okay you want to choose a class monitor okay how many ways can you choose a class monitor eleven right so now how do you read this task your task is to make a monitor so now you will think okay I can make a monitor if I choose a boy correct that's case one or you could also choose a monitor if you choose a girl now both of them are independently helping you to choose a monitor isn't it so the class monitor could be either a boy job is over or a class monitor could be a girl job is over so both of the choices would help you accomplish that task so just find out how many ways can you choose a boy there are five boys so five ways to choose a boy how many ways can you choose a girl six ways to choose a girl because there are six girls in the class so five plus six eleven so the number of ways in which you can choose a monitor is eleven so you add in okay is it fine any doubt related to what is the addition principle of counting that so whenever you use the word cases whenever you use the word this or this or this or this so they are all indicative that you need to add that is why I always encourage students to speak the task in their mind when they speak the task they automatically start using the word or an ad and the moment they do that they know whether they have to add or whether they have to multiply so multiplication is the next thing that we are going to talk about so multiplication principle of counting again let me take an example for the same let us say the same cinema theater okay this time this time I give you a task that could you go back what do you want to copy here anything that you want to copy then okay yeah so let's say now the same cinema theater has got three entry doors okay so these are entry doors and let's say five exit doors five exit doors okay now there is a task which I'm giving you you need to enter the cinema theater okay so what's your task task is to enter the cinema theater entry and exit okay entry and exit and exit what is very important how many ways can you do it right absolutely right so you'll say so three ways to enter and five ways to exit and you eventually ended up multiplying it was was there any justification for it the justification is for every entry you have five exits right so entry number one you can take five exits and in number two five and in number three five so basically you'll end up getting 15 such you know answer and you get it by multiplication now here let's generalize it what was happening here so there was a task and this task was made up of several subtasks okay and in order to consider the task to be completed you have to complete all the subtasks let's say there are k subtasks okay so in this case there was only two subtasks entry and exit right so if you have a task which is made up of let's say k subtasks and you have to perform all of them to consider the task to be done even if you miss one of them the task will be incomplete right so then this principle says the number of ways to do the task so let me define here let's say a task is accomplished let me write it down if a task is accomplished by completing by completing all the subtasks all the subtasks associated with it then the total number of the total number of doing the task or completing the task the task is equal to the product of the product of the number of ways of doing the subtasks okay so here the subtasks associated with this particular task was entry and exit entry is a subtask exit is a subtask and both of them have to be completed then only your task of entering and coming out will be completed so you calculated how many ways to do the subtask of entry three ways how many ways to do the subtask of exit five ways and means multiplication you multiplied and then you get what your answer isn't it now again this looks very simple when I explain it in this example but trust me many people really get confused when to add and when to take multiplication okay so we'll do some initial questions to understand this concept and let me also tell you dear students that these are the formative part of your understanding of counting npr ncr etc the formula that you have already learned in your school they all come from these fundamental principles right so all your concepts are built on this so if these are shaky then those formulas are not going to save you okay so these are the foundation so understand this don't be like don't be in a search of a formula to solve all the questions instead of search for what principle or what logic are they connected with and what principle I can apply to solve those questions we'll take examples don't worry say you're done can I go to the next slide done okay my battery laptop battery is discharging once again let's begin with a question how many different signals can be given using any number of flags from four flags of different colors okay remember you need at least one flag to make a signal so consider yourself to be a station master okay and you know planes are flying so you need to make signals out of certain combination of flags okay and you can use any number of flags but of course minimum is one you have to use one to make a signal and you've been provided with four colors how many different signals can you generate from those four colors provided to you okay sharduli please note that if you alter the positions of the flags it'll count as a different signal let's say you had a flag of red color yellow color blue color and green color okay so let's say you decided to make a two color flag so red up yellow below if you interchange their positions let's say white yellow up and red below it will be counted as a different signal all together no sharduli take no change your answer if you want to very good setu excellent okay tejaswini kartik siddharth nice good shayana very good arav good again good to see you you know participating good doesn't mean your answer is correct so just i'm happy to see you participating yeah anybody else who wants to participate okay sharduli has come back with a different answer pratej very good manu anybody else who would like to answer tejaswini wants to change okay good you can make a signal with four flags also why not noel very good okay let's discuss this see i feel that you have four cases to accomplish your task you can either make a signal with single flag so that's a case one okay case one or you could make a signal with two flags okay second case or you could make a signal with three flags okay looks like flag of nepal right nepali flag is triangular right is there any country who which has triangular flags i mean if you know any nepal for sure has triangular flags okay prism so prism also has changed her answer so now these are the four ways in which your task can be independently accomplished is it so one flag signal is considered to be a signal why not job is done i have to make a signal no job is done but i have several other ways to do it so two flag signal three flag signal four flag signal okay now let's count how many ways can you make a single flag signal you will say sir four options only i could make a red color flag or yellow or a blue or a green four color okay so four colors are four ways to make a single color flag okay how many ways can you make a two flag signal now you say sir if i choose this in four ways then this i can only choose in three ways because one flag is already used up in the first one now will you add these four and three or will you multiply this four and three multiply why you will multiply because fundamental principle of multiplication says if your task is made up of two sub tasks and both of them have to be done because you are making a two signal flag so the first flag and the second flag both have to be put if you put one your job is not over right you have to put both of so by fundamental principle you need to multiply there so this answer will be 12 sorry four into three same way can i say this will be four three two and by fundamental principle it will be multiplied and the other guy will be four three two one which you actually can call four factorial also okay so now what will i do with these four numbers which have circled should i add them or should i multiply them add them right because independently you are accomplishing the task okay so the answer will be four plus 12 plus 24 plus 24 and that answer is 64 ways you can make a signal so this station master can generate 64 signals just by those four flags sufficient for 64 events maybe you know just to show that a train is coming from the other side or maybe some you know a child or an aged person is trying to cross anything i mean just giving examples is it fine any questions great can i move on to the next one let's take this one uh please don't worry about the question number starting from 23 just think this is one two three question all right so the first question i'll put the poll on for this let's read the question first four prizes are distributed among five students the 23rd question says if no student gets more than one price find the number of ways in which this distribution can be done just to answer 23rd question i'm putting the poll on i hope you can all see the poll on your screen so you told change in the color also is a signal yeah i told that so what happened does it change the answer or pradesh no no need to do into why will you do to when you said there are four ways to choose the first flag even if you're changing it that four also accomplish the other change right let's say blue is coming first and then red then red is coming then blue so blue and red are coming in the first position that is counted under four only no why will you do into two okay oh wow i've only got eight eight responses less than a minute good good good people are fast actually this chapter is a touch and go chapter i mean you don't have you know a lot of solving inside but yes initial thinking initial phrasing of your scenario in your mind is important okay i'll give you uh 30 more seconds and then we'll discuss it so people who are answering or want to answer please do so within 30 more seconds okay five oh three two one go okay even if you want to take an intelligent guess please go for it 10 of you have not responded fast fast fast fast fast i'm closing the poll okay most of you have said option number b let's discuss it out see what are the scenario the scenario is you have to distribute four prizes among five students and no student gets more than one price okay now what is your task task is to give away the prizes so you have price p1 you have price p2 you have price p3 and you have price p4 how many ways can you give off price p1 how many ways can you give off price p1 five ways correct how many ways can you give price p2 please keep in mind that no student should get more than one price now the person who has got price p1 he is not eligible to take p2 right one price is you know not not counted in our you know eligibility for p2 right so p2 will have only four options right similarly p3 will only have three options that means the people who have other persons the students who have received price p1 and p2 they are not eligible to receive price p3 similarly p4 can be received by only two students so now what should i do add them or multiply them sir if you add them none of the options will match okay that's a smart way to solve the problem we'll multiply it why because i have to give price p1 and p2 and p3 and p4 see i'm always using this connector and and and and and means multiplication so this answer is going to become 120 option number b is definitely right well answered okay simple easy thing yeah i mean this is not a rocket science you know you just have to use your understanding break it up into tasks and see how the the work can be done the task can be done okay next one i'm again relaunching my poll the next question says if each student is eligible for all the prices find the number of ways in which this task can be done and this is the simplest question oh wow three people have already answered four have already five have already answered good and all of them have given me the same response good i think we should we can close this within one minute that's such a easy question writes it does okay last 20 seconds those who want to vote please do so oh three two one go oh oh ah it was supposed to be 100% vote for option a but unfortunately two students voted for b and c okay so out of 16 14 say a 11 students say b and c maybe out of desperation you have pressed on some whole button see very simple again my task is what my task is to give away these four prizes p1 and p2 and p3 and p4 how many ways can i give off price p1 you'll say five how many ways i can give off p2 again five because the student who has received p1 he's also eligible to receive p2 because as the question says each student is eligible to receive all the prizes okay so five into five into five because you have to give off all the prizes so five to the power four that's nothing but 625 option number a is correct can we now do the 25th one as well let me launch the poll for that as well read the question carefully question says if no student gets all the prizes find the number of ways find the number of ways in which these prizes can be distributed good good good so seven of you have voted last 30 seconds five four three two one go please vote please vote enough time has been given for this okay great so 14 of you have voted out of which 10 of you say option number b which is 620 let's check whether 620 is correct or not see here the question says that there should not be any case where all the prizes are backed by one student okay so can i say there are only five such cases okay where all the student where all the prizes go to a single student right let's say student number one bagging all the four prizes that's one case so number two bagging all the prizes second case so number three bagging all the prizes third case so number four bagging all the prizes fourth case and so number five bagging all the prizes that is the fifth case right now from the total number of cases that you have that is there is no restriction at all that means any number of prizes can be given to any number of students if you remove these five cases so 625 was a case where there was no restriction from those cases if you remove the five such cases where all the prizes go to one particular student you will be left with such cases where not all the prizes go to a single student that means a student can get at the max three prizes not all the four okay so this is a way which we solve which we use and this is called the exclusion way of solving the problem so instead of saying instead of doing all permutation combination let's figure out in how many ways can the task be actually not done that means if you want to give all the prizes right how many ways can you do it so there are only five ways in which you can give all the prizes to a single student those five ways just remove them remove them from what total number of cases which is 625 so the remaining case which is 620 is the number of ways where no student will get all the prizes right but still there are cases here where a student can get more than one prize also but he will not get all the prize that is for sure is it fine any questions so this is how I want you to start you know thinking when you're trying to solve a question you tell me if there is another way to do it and how difficult will that maybe so let's say if you want to do this question okay you have to take a lot of cases number one let's say sweet number one two three four five okay you have to start distributing the number of prizes in such a way that no number is five and the sum is five are you getting my point so you start breaking five as four such numbers none of them being five such that they add up to give you a five something like zero one two two zero okay that means student number sorry you have to break into five so there so you have four prizes right they are four prizes right so let's say student one student two student three student four student five now see the complexity of the way if you want to do it in another way let's say four prizes you're distributing so let's say this gets this guy gets zero this guy gets one this guy gets one this guy gets two this guy gets zero right now here also you need to understand which prize which of the four prizes this guy has got so there is four c one way to choose this right then there is three c one way to choose this then this is two c two ways to choose this so this is one possibility there could be so many other possibilities like one one one zero two sorry there are only four prizes right this is also one okay so there could be so much so many cases so multiple cases will come up and you will you know go mad counting them so that is the only smarter way to solve the problem other ways would be leading to complicated counting our purpose is not to complicate our counting our purpose is to simplify our counting okay all right so let's take more questions let's read this to this question the number of ways in which five letters be mailed in three different mailboxes if each letter can be made in any mailbox let's put the poll also on question understood there are five letters and there are three mailboxes okay how many ways can you post these letters in these three mailboxes okay sharduli siddharth good good you have given me a response personally also that's good no problem sharduli situ if you want to type your answer in the chat you can go ahead most welcome to do that i could see 14 of you have voted so far i'll give 30 seconds more okay five three two one okay now you can see in your poll result i'm sharing this result with everybody now you can see equal number of votes have gone to b and d you know why because people are confused whether the answer will be three to the power five or five to the power three and i know this confusion was evident okay again this is where your basic you know breaking of the task comes into picture so what what are the what are the tasks that you're doing here posting letters correct how many letters are there five letters l1 l2 l3 l4 l5 okay so your task is to post letter number one and letter number two and letter number three and letter number four and letter number five how many ways can you post letter number one write down on the chat box three ways three mailboxes are there how many ways can you post letter number two that also three ways letter number three that also three that's also three this also three okay or what happened to my three three okay so your task is to post all the letters let letter l1 and l2 and l3 and l4 and l5 right so your task will not be completed till you post all the letters so the answer to this by fundamental principle of multiplication will be three to the power five option b is correct is it clear is it clear any questions any questions now see if you are going by you know counting which letters go into a mailbox they just finished it will become our tedious process of counting because all five letters can go into one mailbox that doesn't mean you are you are you know putting all the see you can choose your letters which letter will go into which mailbox right so there would be a complicated distribution just like the prices and the students prices are like let's say you know you can you can you know relate it to any of the two so which price will go to which student that also will become complicated so instead take the work as posting the letters and each letter have five options to be posted that would be a wiser way to solve the question or a smarter way to solve the question are you getting my point five to the power three will not be the answer the reason being see what are you trying to say you're trying to say that a let a mailbox has got five options it's not about five options there a letter letter has got five three options to fall in any of the letter box because the mailbox doesn't know how many letters are going to fall and which letters are going to fall in that but let us know that there are only three options for me I can use any one of the option to get inside right so please understand this if you have not broken up your task wisely you will end up getting wrong results okay now let me give you a synonymous question let us say in a lift let's say in a lift okay this is a lift okay this lift is carrying eight people okay I'm just making eight people in this lift big lift okay okay so these are the lifts okay this lift stops at five flows so there are five flows in which this lift stops okay floor one floor two floor three fourth floor and fifth floor okay in how many ways can these eight people alight or get down on these flows given that any number of people can get down on a floor so in how many ways can these let me write down the question in how many ways in how many ways can the eight people or can the eight persons alight on these flows on these on the five flows these flows where any number of people can get down on any floor again I'm getting two answers eight to the power five five to the power eight what is your task is your task to fill the flows as a task to make the people get down ask your question ask that question to yourself is was your task in the previous question to fill the mailboxes or to post the letters very deep question is this okay if you're not able to answer this you'll not get a right answer so what was your task in the previous problem posting the letters or filling the mailboxes posting the letters what is your task here making the people get down or make or you know filling the floors with people making the people get down right so if you're making the people get down person number p1 should also get down p2 should also get down p3 should also get down till p8 should also get down correct so now you are providing options to these people so person number one how many options you have to get down you will say I have five options to get down person number two you're asking him how many options you have to get down he'll say I also have five options to get down five five five what will you add what will you do to these five five five will you add them or will you multiply them add them or multiply them multiply them because all the people should get down in order to accomplish the task the answer to this question is five to the power eight not eight to the power five understood any questions any concerns okay let's take more interesting questions I mean I can keep on giving you more questions so let's say let's say there is a room okay this room has got 10 light bulbs in it okay so there are 10 light bulbs of course I'm not going to make all the bulbs dot dot dot okay there are 10 light bulbs in this room okay and these bulbs are operated by independent switches okay so this is a switch for each of these bulbs okay switch one switch two switch three switch 10 okay this is bulb one bulb two bulb three okay my question to you is in how many ways can the can the room be lightened up in how many ways can the room be lightened lighted means there should be light in that room okay okay shardhuli very good nick hill very good anybody else only two people effectively have answered this question what about others okay vashtav okay karthik so karthik if you light two bulbs will the room be not lightened if you light three bulbs will not be room lightened you further reduce your answer okay now I'm getting weight varieties of answers right from 10 factorial 100 190 of course there are some right answers also in this I will not tell them actually now most of you are just trying to force fit factorial right just because you have learned factorial use let's use it no okay so see understand here logically when will the room be lightened when at least one bulb is lightened it doesn't matter how many but at least one bulb should be lighted now if I ask you this question how many ways can you operate on these switches what is the total number of ways you can operate on the switches now remember a switch has got only two states on or off okay so let's say I write down the options that I have for all my switches okay so every switch has got two options whether to be on or whether to be off right so you can on the switch one or off the switch one switch two also on off switch three also on or off on or off on it can't have both the states together it can only be in one of the states okay so all together how many states you can provide to these switches and you have to provide one of the states either it could be on or off it cannot be in a third position right so all together you can provide two to the part 10 states to these switches that means 1024 right now in this 1024 there is one such case where everybody will be off isn't it and that's the only situation when the room will be dark remember what did I say at least one switch should be on for your room to be lighted up isn't it so there is one situation where the room will be completely dark when all the switches have been in the off position that one is something which you need to exclude from your answer so the room will remain dark room remains dark only in one way dark sorry I wrote a dark okay in only one way so this you exclude from here so if you exclude that one way from here 1023 ways or two to the part 10 minus one way is the ways in which the room can be lightened up this is your answer to this question people who are saying 100 people who are saying 10 people who are saying one or people who are saying 10 factorial basically you had to use 10 somewhere that is why you were giving those answers okay so it does when you're saying so when you're saying you want to make the switch like a room lightened up you can switch on any number of combination of these switches or any number of combination of these bugs right so what I did first I I found out all possible ways I can switch on or switch off these bugs so that is 10 2 to the part 10 and in that 2 to the part 10 there's only one way where all the switches will be off everything that's one way right that one you subtract so when you subtract one you are basically ensuring that at least one switch is on and because of that room will be lightened up same amount of in how many ways will the room have the same amount of brightness no 10 will not be the way see same amount of brightness you're talking about brightness of one bulb so if there's a brightness of one bulb there are 10 ways in which you can brighten above by one bulb if you want a brightness of two bulbs that normally brightness is measured in terms of lumens or in tv we call it as nits okay so two bulb brightness you can choose 45 ways to do it 10 c2 any two bulb you pick up and on it okay three bulb brightness 10 c3 okay are you getting my mind so that that answer will be something else okay that answer will become something else okay so basically no identity will matter no because every person is a different person let's say out of those eight people no person number one getting on floor number two and person number five getting on for number two will be counted as a different combination right people are all distinct no they have an identity okay so i have a question for you all here let's take this question and then we'll take a break after this the number of ordered pairs m comma n where m and n come from this set of one to 50 such that 6 to the power m plus 9 to the power n is a multiple of 5 and again i'm putting a poll on for you all here i'll give you sufficient amount of time to work this out okay i think two and a half three minutes is good enough yeah oh i think one of you was in a rush to answer this so i've already got one response in almost 11 seconds and that is wrong by the way yes yes there is only m and n can be same numbers also right last minute oh i could just get two responses so far okay down on begins five three okay two two minutes more i cannot give because see the lifespan of a question is only four minutes say to understand this and we need to keep in our practice also if time is not considered then j advance paper is also easy right give me one month i will solve j advance paper okay it's not one month you have to do it in three hours right if you remove the time factor form up in a question then the question is a dead question i can give one more minute if you want okay shardily i can see i'm very mixed response coming from the crowd all the options have been touched upon okay so let's let's discuss it i'll close the poll now okay anybody wants to vote take a guess if you know if you're not if you couldn't solve it take a guess take an indian guess when i say take guess people normally go for b option but don't do that here be choosing b would be hazardous okay now most of you have gone with c by the way c is the most chosen option let's discuss it out see when you talk about when you talk about just give me a second yeah so when you talk about this type of term six to the power m okay and nine to the power n okay now see here if you want to find out how many you know numbers or how many combination of m and n are possible for which this is a multiple of five okay a multiple of five has a very important characteristic which you everybody knows here that it should end with a five or end with a zero correct so any multiple of five any multiple of five okay it should either end with a zero or end with a five yes or no now if i raise any natural number power to six not only limited to one to fifty any number you raise this will always end with a six or end in a six isn't it six to the power one ends with a six six to the power two thirty six ends with a six six to the power three two hundred and sixteen ends with a six okay and any multiple of nine either ends let me write with only because i have used this ends in end with both are fine no issues let's not worry about the english in one day nine to the power n and of course n could be any natural number could either end with a one or end with a nine itself nine to the power one ends with nine nine to the power two ends with one nine to the power three again ends with nine nine to the power four again ends with one so either end with one or anything okay now in order to know whether the sum of this number okay what will it end with it will either end with a seven or it will end with 5 because 9 and 6 will add up this will end with a 5. And this will end with a 7. Now I want those cases where they end with a 5 because only the ones which end with a 5 will be multiple of 5. This is rule out, this is the one which we desire. Now in order to end with a 5 is there any restriction on m value? Because you can't place any restriction because any number you choose from 1 to 50 it will end with a 6 only. So there are 50 ways to choose m. But for 9 to the power n to end with a 9 you can only put n value as 1, 3, 5 that is nothing but yes all the odd numbers you can put. So there are only 25 ways to choose your n. So there's no restriction on m you can choose any number from 1 to 50 that is fine. So 50 ways to choose your m but 9 to the power n if you want to end it with a 9 and hence the sum should be a multiple of 5 you can you only have 25 ways to choose your n that is all the odd powers from 1 to 50. So your task was what? Your task was choosing your m and n both, this and this both. So this you can choose in 50 ways, this you can choose in 25 ways and means multiplication answer is 1250 ways right. So option number a was correct sorry option number a was correct. Let me see the poor result only 4 people got this right as per the poor result out of 20. Is it clear any questions any concerns please highlight. Okay. Okay. No issues. Fine. Here we'll go for a break. Okay. The time as per my watch is 613. I will see you at 628 p.m. exactly after 15 minutes break. Eat something, rejuvenate yourself and we'll meet in 15 minutes time. Okay. See you on the other side of the break. Start. I'm going to start with your understanding of the permutations concept. Okay. So officially the first topic that we are supposed to talk about in this chapter now has begun after so much of groundwork being done, especially knowing your fundamental principles of counting because without them you will not be able to understand what's permutation, what's combination and how to accomplish these countings. Okay. So as I've only discussed with you permutation basically means arrangement and arrangement means you're not only selecting items that you want to arrange, but you also are giving some order to them. Okay. So here the very first thing that you would come across is the constant use of this symbol NPR. I'm sure many of you would have already seen it in school. By the way, in some books they will write it as P with argument of NNR. Basically they want to give it as a look of a function just like a function has got some kind of inputs. So here P is the function which obviously manipulates or works on these inputs and gives you some result. So NPR is basically nothing but I mean let's find out the way to read this formula also or this symbol also. NPR wherever you see it's basically a symbol to say number of ways to arrange, arrange our objects taken from N distinct objects. Are you getting my point? So this word distinct is very important. So if the objects, these N objects are not distinct that means if some of the objects are identical, you cannot use NPR or the symbol will not be utilized in solving that kind of a question. So if at all you are using or you're trying to give an answer to that question or if at all you're reading that there is an NPR, the first thought that should come to your mind is NPR means number of ways to arrange our objects. Okay, of course your R has to be less than equal to N. So number of ways to arrange our objects taken from N distinct objects. Are you getting my point? This is very important. So for example, if I say how many ways can you arrange two objects taken from three objects, you will say 3P2 but only when these three objects are distinct objects that means they should not be identical. So A, B, C if you have three distinct objects, how many ways can you take two objects and start arranging them or ordering them? So you can do it in the following 3P2 ways. Okay, we'll come to that final value of it but this is how what 3P2 value will be. In this case, it will be AB, BA, AC, CA, I'm sorry then BC, CB. Okay, later on we'll learn how 3P2 value is actually 6. Is it clear? If you're any two objects were identical, let us say it was AAB and I said take two objects and arrange them. Now remember here if you take A and A, there's no point writing it again A and A both will be counted as the same. So either you have A, either you have AB or you have BA. So there are only three ways to do this. Okay, so this is not 3P2. This is not 3P2. This is 3P2. So 6 is your 3P2. Okay, understand this very, very important. Now symbol you have understood, you have learned the meaning of the symbol. Now let us derive the result for the symbol. What is this expression equal to? So in terms of NNR or in terms of factorial, etc., whatever you will be coming across in some time, what is NPR equal to? What will NPR give us? Okay, so we'll start from scratch. We'll start from fundamental principle to derive NPR. So NPR is what number of ways of arranging R objects taken from N objects? Correct. So think as if there are N students. By the way, students are identical or students are distinct objects. Every person is distinct, isn't it? Right? Each student is different from the other. Yes. Now let's say there are N students and there are R chairs. Okay. There are R chairs. Okay, so there are R chairs. You want to arrange R students taken from N students over these R chairs. So you're making a linear or a row wise permutation of R students taken from N students. How will I do this question? How will I accomplish this task? Now one way is you choose them and then you order them. But again, choosing something is not something which we have done. We'll do it later on. Combination is what we are going to take up later on. So let's say irrespective of whether you know or you don't know, how will you start filling up these places? How will you start filling up these chairs? Let's try to answer it. First chair, how many options you have to make a person seated on a first chair? Or how many contestants do you have for the first chair? R, R contestants. N people are there, no in front of you, Karthik. You have N people contestant, right? Anyone of them can be, is eligible to sit on this chair, right? Okay, great. How many contestants or how many people are eligible or how many options you have to fill up the second chair? N minus 1 because one person has already been seated on first chair, okay? How many options you have for the third chair? You'll say N minus 2. Fourth chair, you'll say N minus 3. Okay, now think carefully and answer. How many options you have to fill the R-th chair? Okay, okay, Nikhil, connect, Situ, Tejasini, connect. Okay, all of you have given the right answer. So it's N minus R plus 1. Okay, Karthik, both Karthiks. See, first chair the pattern, Karthik. First chair, N minus 0 is, that's N actually, right? Second chair, N minus 1. Okay, so whatever is this number, subtract one less than that number from N, that is your answer. So it's actually N minus R minus 1, which I've expanded to N minus R plus 1. That's why I said think and answer. Anyways, now as per fundamental principle of counting, will you add them or will you multiply them? Will you multiply, you'll multiply them exactly because you have to fill the first chair and the second chair and the third chair and the fourth chair and so on till the end of chair. So as per fundamental principle of counting, you will multiply these results and there you go. This expression is your formula for NPR. Got any questions, any concerns? Now, many people say, sir, but our teacher told us in terms of factorial, etc. No problem, we will simplify this in terms of factorial. I repeat what part Manu, the whole thing, the whole process. See, we are trying to find out what is NPR expression, right? NPR expression is number of ways to arrange R people chosen from N people in a row wise or in a chair. So I've taken a case study for this. Let's say there are N students and you have R chairs. How many ways can you arrange R people on these R chairs taken from N people? So for that, I started the process by filling up each chair one by one. So first chair can be filled in N ways, right? Do you agree with me on that? Isn't it? So let's say there are 10 students, you have three chairs. So first chair, you can make any one of the 10 guys sit. So second chair, any one of the nine guys sit. Third chair, any one of the seven guys, eight guys sit. Same way, I'm basically filling up these chairs. First chair, N ways. Second chair, N minus one way you can fill it. Third chair, N minus two and so on and so forth. So R-th chair can be filled in N minus R plus one-th way. Let me give you a simpler number to understand. 10 people. Okay. First chair you can fill in how many ways? And we have three chairs. First chair, how many ways? 10 ways. Correct? Second chair, nine way. Third chair, eight way. Isn't this 8, 10 minus 3 plus 1? That's why N minus R plus. Clear now. Sir, if all the students can sit in any one of the R chairs, both the second and third, etc., also be N in. No, no, no, no, no, no, no, no. See, you are basically making people sit, right? And once the first chair is occupied, it is occupied by some person, right? Remember flag question. Once a flag, first flag is chosen, you can't choose the same color for the second flag. No. So how can one person occupy two different chairs at the same time? Okay. Okay. Now, this expression that you have over here is a slightly complicated looking expression. So many a times in the books and in the school, they will simplify this expression. How do they simplify this expression? All of you see, I'll rewrite this expression once again. Let's say I multiply with the term which is coming exactly after N minus R plus 1, which is actually N minus R. And I also divide with this term. Will it change my expression? Does the original formula change by multiplying and dividing with N minus R? All right. Yes or no? Does this answer change by multiplying and dividing the expression by N minus R? No. Okay. Let's continue doing this activity. Let's now multiply and divide with N minus R minus 1, N minus R minus 2. And let's say I go all the way down to 1. This also all the way down to 1. Do you see that the numerator is actually a continued product from N till you reach 1? What is the continued product of N till 1 called in language of factorial? N factorial. What is the continued product from N minus R to 1 called N minus R whole factorial? Right. So this is an alternative and easier looking formula which you can keep in mind for remembering your NPR expression. So NPR is N factorial by N minus R factorial. Now, remember a question I gave you a little while ago where you had to make arrangements of two letters chosen from ABC. So the answer was 3P2. So in expanded version, it is 3 factorial by 3 minus 2 factorial, which is 6 by 1, which is 6 ways. That's why the count came out to be 6. Is it clear? Any question, any concerns related to NPR here? Okay. Let's look into some properties of NPR. If this is done, we can move on to the next slide and we can look at the properties of NPR. Any question, any concerns, anybody? Do let me know. So you can see this is not a rocket science formula. This formula was not given to us by any saint or any god. It came from a logical expression of fundamental principle of counting. I have a snapshot of certain properties which I would like to take up one by one. Okay. Okay. By the way, you can just ignore this fourth one. I'm just scoring it off. So let's talk about properties of NPR. Just to save time, I've taken a snapshot. Look at the first property. It says that if you want to arrange N different things taken all at a time, it will be NPN, which is actually correct because out of N things, if you take all the N things and start ordering them or in short, you are arranging N things among themselves, then the number of ways to do it as per the formula is NPN and that comes out to be from your formula perspective, N factorial by N minus N factorial, which is N factorial by zero factorial, which is actually N factorial itself. Okay. So remember what did I give you when I was talking about the definition of a factorial? Factorial represents the number of ways in which you can order that many substances which are distinct first of all. That's why this word different becomes important. So N different things among themselves means taken all at a time. Okay. So formula also of NPN says it is going to give you N factorial only. Okay. Apart from this, there are some special values that you can keep in mind like NP0 is one. NP1 is actually N. Okay. Now many people say, sir, how is NP01? Because if you read this, it basically sounds very weird to us. How many ways can you arrange our objects taken from N distinct objects? Doesn't it sound weird? It does, right? Yes or no? Okay. So when you say how many ways can you arrange zero objects taken from N objects? So if you're not taking any object, how can an element happen? That thought comes to everybody's mind, isn't it? It came to me also when I was preparing for my cognitive exam. But again, this is a mathematical inconsistency. And in order to solve this, in order to solve this mathematical inconsistency, we basically rely on that formula or rely on this fact that there's only one way. That means nothing you can do. When you're not choosing anything, how can you arrange them? Right? So we call that only one way. Okay. So there's one way of choosing nothing and arranging them among themselves. Okay. You can't say zero also because that will create a mathematical inconsistency. That is why one. Is this fine? Any questions? Okay. Now coming to the third, from third onwards, we should be slightly, you know, serious in these properties because they're very useful. If you see the third property, it says NPR can actually be broken down into N, N minus one PR minus one. Okay. Now let us try to logically prove this. What is a logical proof for this? Can anybody logically prove this? Logically means with logic, not with formula. Formula, I know you can prove it very easily. Just like how you basically played with your fundamental principle of counting, can you basically prove this also? Just by logic, no formula, logic. Okay. Let me, let me give you a situation here. See again, let us say there are R chairs. Okay. There are R chairs and there are N students. I want to arrange them, I want to arrange R students taken from these N students over these R chairs. Or in short, I want to arrange R people taken from N people. Okay. So one way to do it is NPR. Another way to do it is, this is one way to do it. Another way to do it is, let's say I break this event like this. One of the students here, let me name somebody, maybe say two. He says, sir, I know, I'm feeling very hungry. I'll come back in five minutes. I'll have something from my school canteen and I'll come back in five minutes. Meanwhile, you arrange those, the remaining people on those chairs. So let's just say two has left the scene. How many people are there in the crowd? How many people are there in the crowd? N minus one. Oh, sorry. Let's make a different story because I have a different thing to prove. Okay. Let me do it in this way. Now, let's say, sorry, I'm cancelling it out because that is not applicable because I did not see this part of the question. Yeah. See, let's say one chair is a special chair. Okay. Any chair you choose and call it as a special chair. Let's say this chair is a special chair. You can say the Maharaja chair. Okay. Now, this chair I want to fill first because that is a special chair, a chair of authority. And I want to fill that chair first. How many ways are there to fill this chair? Anyways, correct? So I fill this chair first and and means multiplication as per fundamental principle of counting. And I then fill the remaining R minus one chair from people chosen from N minus one left of people because one person has already come and sat on this chair. Correct? So remaining N minus one people are still there to be chosen from. And I have to fill the remaining R minus one chairs from those N minus one people. Correct? So that task can be done in N minus one PR minus one. So eventually, what am I doing? I'm filling all the R chairs from people taken from N people, isn't it? Whether I do it by this method or whether I break the task like this and do it. Eventually both the countings are going to be equal. That means your NPR and N into N minus one PR minus one would be equal. Correct? Yes or no? See, why I'm giving a logical proof is because it will, you know, start making you think how to solve a problem in permutation combination. If you start thinking like this, then you'll feel that this chapter is very easy. Right? You can work your way out in any difficult situation. Else you'll always start thinking in terms of some formulae, which is not going to help you out in the longer run. Okay? Now if you see the formula also says N into N minus one into N minus two PR minus two. Can you prove that also in a similar way? Exactly. That's what has been done, Karthik. Now here you could also do this work in another way. You can say, okay, let two chairs be special chairs now. Right? Exactly. Very good. So let two chairs be special chair. So let me fill special chair number one first and then I'll put special chair number two. So one can be filled in N ways, two can be filled in N minus one way. The remaining R minus two chair has to be filled from people taken from N minus two left of people. So this is chair number one, special chair number one, special chair number two, and the remaining chairs, sorry, N minus two PR minus two. Okay? So your task can also be done by, you know, doing these multiplications. This will also give you the same answer. And you can continue, you know, proceeding on and on till you basically have, till you basically have exhausted all the numbers. That means all your chairs have become your special chairs now. Okay. So this is how I would like you to start thinking or at least inculcate that thought process slowly and slowly as you are, as you are learning this chapter. Okay. Now many people ask me, sir, these properties that you have written, that is number third property or the properties which are going to come henceforth, do we need to remember those properties? No. Don't remember it. Understand it. Right? Let's do the fifth one. Why I scored off the fourth one is because fifth and fourth are actually the same properties. Okay. Yes, fifth one. Fifth one says that NPR can actually be written as N minus R plus one into NPR minus one. Well, what is the logical proof for this? What is the logical proof? Can somebody give me a story and convince me that, sir, see, if you want to arrange our people, sorry, you want to take our people from N people and arrange them on our chairs, it is the same way as doing this activity or doing this, you can say operation. Very simple. This is actually the reverse of what we did to prove the previous one. Hint. Who will tell me? Take N chairs. Chair, okay. Then at the R-H chair, it will be, okay, take R chair. R chair will be N minus R plus one. So remaining chair, why N minus R plus one? See, now get this right. I say, so don't get confused. You are trying to mix the previous proof with this. Sorry, the derivation of NPR with this. See again, let's say there are chairs, okay. And one chair is a special chair, but this special chair you want to fill at the last, okay. We'll fill this at the last. So first what I'm doing, I'm filling the remaining R minus one chairs, right? So number of people in the crowd is still N and I have to arrange R minus one people taken from those N people and put them on the chairs, correct? Now, once all these chairs are occupied, I have to fill those special chairs. So that is the last thing that I'm going to fill up. Now tell me how many people are waiting in the crowd after R minus one people have sat? How many people left? Tell me. All right. N minus R minus one people are left. So that many options you have for the special chair. So this into this is as good as NPR. Done. This is the proof. If you take our chairs, remove one of the chair, then the chair left is R minus one. So the number of people required to fill the rest of the chair is N minus R plus one. I didn't get that Karthik. If you remove one chair, there is R minus one chair, okay? And you want to arrange R minus one people on those chairs chosen from N people. So that's NPR minus one. But that doesn't match with NPR, right? Because NPR is arranging R people from N people, but you only arrange R minus one. How will that both match? That one chair you have removed that you are talking about? Oh, that is what I did indirectly. That is what I did here also. How is N minus? N people were already there. You filled the yellow tick mark chairs. How many people have already sat on the chair now? This many people have sat. This chair is vacant as of now, correct? So this many people are left to be seated, no? Tejaswini? I mean, don't ask the silly questions here, at least basic questions. R minus one people are gone. They have seated on the chair. How many remain? N minus R minus one. What is there to ask there? Okay. Is it fine? Okay. Now, I will basically, it's not end of the properties. I'll give you some problematic statements. Okay. And then I'll ask you to get to those results. Yeah. Let's take this. Okay. Now, this is a property, but I have written it in a form of a question. Okay. Question is prove that number of permutation of N different things taken are at a time when a particular thing is always included in each arrangement is given by R into N minus one PR minus one. Read the question. The question is basically having a restriction into it. The restriction is there is one particular thing, which is always a part of your arrangement. Prove that the number of arrangements possible is R into N minus one PR minus one. Again, logically think this out. You will be able to get your answer. So in order to prove this again, I will use a logic. See, there is a particular thing, which is always to be there. Now this thing is already given to me that, okay, so this is the thing which you always have to put in your arrangement. So what I'm going to do, this N thing I'm going to break up as, let's say I take it as a situation of students. Let's say there is N, there are N students and there is one special student there. Okay, let's say that special student is Mr. A. Now out of those N students, there is one student which is special called Mr. A. A could be any name by the way. And there are N minus one other students. Okay. Now I have to make an arrangement of R students taken from those N students where Mr. A must definitely be there. So let's say I want to arrange them on these R chairs and Mr. A has to be there in this arrangement. Okay, he's a special guy. He's, let's say a special person. He has to be there. So what do I do here? I ask Mr. A, hey, go and pick up a chair where you want to sit because you're definitely there in the arrangement. So tell me how many options will Mr. A have to choose a chair, right Tejasini, Light Kartik. So A can choose a chair. So A can choose a chair, a chair in our ways. Okay, in our ways. So he's that, you know, special guy VIP treatment is given to him, right? So he says, Oh, go boss and choose your chair where you want to sit. You're the VIP person. So this fellow goes inside and he basically takes, you know, careful observation and he picks up one chair and goes and sits over it. So he has got our options to choose from. And the remaining N minus one people have to be arranged in how many chairs remaining. So one chair has already been occupied by A. So R minus one more chair remain and N minus one people are there. So these N minus one people. So R minus one people taken from N minus one people or students can be arranged in can be arranged in N minus one PR minus one way. Now, will your task be complete? Will your task of arranging our people on these N chair on these chairs will come will be complete if you just do one of them? No, right? You have to do both of them. So Mr. A also has to be selected and you have to ensure that our people are also seated, including Mr. A. So by fundamental principle of counting, you will multiply R with N minus one PR minus one. So this is the number of ways in which you can accomplish this task where that particular thing is always included in your arrangement. Now here, many people say, sir, don't we have to choose that particular thing? No. The word particular means that the thing has already been chosen. You don't have to apply any kind of a choices on that. So if you see the word particular, if you see the words specific, that means that object is already chosen and given to you. You don't have to apply an extra, you know, counting there to choose that particular thing. Are you getting my point? Be very, very particular about this word particular, right? Because it appears in your questions and many people misinterpret it and get a wrong result. Yeah, Karthik, kind of a minister's, let's say some minister's son. So he is given some special preference. So he is the one who is the first one to occupy the chair, right? You can think like that. Is it fine? Any questions here? Okay. Let's take another situation. See, I'm just exposing you to many situations. I mean, my purpose is not to make you mug up any set of formulae. Okay. If you are learning enough to think like that, you can tackle any question that comes your way in this chapter. Okay. Let's take this question. Number of permutations of n different things taken are at a time when a particular thing is never taken into the arrangement. Now, this is even simpler. You say, let's, sir, Mr. A is never taken. So since he's a minister's son, okay, we will never entertain him. Okay, maybe because he's not eligible to be a part of the team. Okay. So here, what will you do? You have effectively n minus one people to choose from because one person has been ignored because he is to be never taken into arrangement. So you have to complete a selection on arrangement of our people. You have to do an arrangement of our. So these are will be basically taken from these n minus one people and arranged. So that will be n minus one PR. Does it even need any kind of a writing from my side? Understood whatever I said. Do I need to write down something to tell you or understood? Understood? Okay. Great. Let me take you to another situation. Okay, let's do this one. Prove that. Prove that the number of permutations of n different things taken all at a time when n specified things. Now, see the word specified is very important. They always come together. Okay. So there are n specified things which are always kept together in that arrangement. So how many ways can you do it? That result is given to you by this expression m factorial into n minus m plus one factorial. Can somebody tell me how does that result come? Anybody? Shardili has figured out the way to do it. Okay. Shardili, can you explain how does it work? How do you get this m factorial and how did this m minus n? Yeah, why not? Please unmute yourself. Anybody can unmute and talk please. Yes. Hello, sir. Am I audible? Yes, Shardili, very much. So they said m specified things should come together all the time. So if you draw n, n hyphens of hyphens, n things. And if you want m things to come together, then you can place those m things in n minus m plus one ways because if it's m from the beginning and the totally n different things, then it's n minus m plus one from the end. So you can have m different sets like that in the whole arrangement. And if you have m sets like, sorry, you can have n minus m plus one sets like that in the whole arrangement. And then those m things can be arranged in m factorial with amongst themselves. So it becomes m factorial into n minus m plus one factor. Absolutely right. Now, see, whatever she has said, I will also explain it to everybody. See, what I normally do, I use a method which I call as the string method. Just to give fine tuning to the method that Shardili has discussed. So I normally use something called string method. What is the string method? So those m specified things which we are keeping together, I tie them with a string. Let us say those m specified things are o one till o m. So what do I do? I take a string and I tie them like this, how you bundle up objects. So the reason I have tied them is just to show you that they become one single unit for you. So whenever they will go, they will go as a unit so that they are always together in the arrangement. Isn't it? So it's like, let's say you have 10 books and three books you want to keep always together. What do you do? You bundle up those three books. Now wherever you take, however you arrange those books, those three will always be together. Isn't it? Now, what Shardili is saying that if let's say this is one object, let's say one unit and these are the remaining n minus m units, then total number of entities or the total number of units that you will have over here is n minus m plus one units. Isn't it? So these will be the total number of entities. So one bundled one and the remaining n minus one together will be one plus n minus one, which is this. Now how many ways can you, how many ways can you arrange these objects taken all at a time? What will you say? How many ways, number of ways to arrange n minus m plus one distinct objects among themselves? So they're all distinct by the way, even though they're all bundled up, that bundled up is counted as one object. So how many ways can you arrange these n minus one distinct objects among themselves? You will say, sir, it is n minus m plus one factorial, right? Anybody has any doubt with this? Anybody has any doubt with this? Okay. And these m objects that you have tied with the string, the number of ways to arrange these m distinct objects among themselves, see, they can be interarrangement also, right? So these m objects, see, they're together, that doesn't mean it is always going to be in this order only, O1, O2, O3 till Om, no. Om can be the first one also. O3 can be the second one also, right? So you can do an intra-shuffling of O1 till Om and that can be done in m factorial ways, correct? So what do you have to do? You have to arrange those objects, keeping those m objects together. So fundamental principle of counting says both of them have to be performed. So as per your fundamental principle of counting, your answer will become this into m factorial, write it before or write it after, that doesn't make any difference. Is it clear? Any questions? Let's take a small instance or question based on this concept. Let us say, there are 10 friends, okay? There are 10 friends, five of them are boys and five of them are girls, okay? So let's say in a classroom, there are five girls, five boys, they're all friends. They want to go and go for a movie together. Let's say the class is going for a movie together, okay? Now in how many ways can these 10 students be seated in a row such that all the girls are seated together? So how many ways, how many ways these five boys and five girls can be arranged in a row such that all the girls are seated together? Let me write it down. Find the number of ways to arrange five boys and five girls in a row such that all the girls are seated together. Again, one thing I would like to all tell you, don't give your answer in a calculated manner, give it in a very raw format so that I can understand the logic behind it. Kartik, good. Good. Kartik has directly used the formula, very good. Manu, okay. Siddharth, good. Anybody else? Okay, Tejasini, very good. Okay. Now if this formula remember well and good, but I would not never encourage you to remember this formula. Try to solve this question from the basics. So what I will do, what I will do, see five boys are there and five girls are there, okay? So what do I do? I take a string and I basically tie them up and treat them as a single person, that's it. So all the five girls are treated as a single unit. So wherever they will move, they will move as a single unit so that they are together. So overall, you have one, two, three, four, five, six, six entities or six units. So that can be arranged in six factorial. But this doesn't account for the fact that the girls can be together even while they are shuffled among themselves. So it is not necessary that your girls row should start with G1 only, right? It could be like G2, G3, G1, G5, G4. So there could be again, five factorial, entire arrangement of the girls, correct? That's it. Answer is six factorial to five factorial, absolutely. Why sharduli that extra five factorial? Oh, sorry, it was a mistake. Okay. Yeah. Is it fine? Any questions? Okay. Now, one more thing I would like to discuss here is this property. Number of permutations, or you can treat this like a question, prove that the number of permutations of n different things taken all at a time, when m specified things never come together. So now again, read it in light of the question which I gave you. Five boys and five girls are to be arranged in a row such that the girls should never come together. That means there should be at least one girl who is away from the group sitting somewhere else. Okay. That means all the five should not come together. How many ways can you do it? Now, if you look at the answer that they have asked you to prove, it basically makes everything clear. You will say, sir, first of all, n factorial is the number of ways to arrange them without any kind of restriction put into place. So this term is when there is no restriction. Okay. That means 10 people you are arranging on a row. You can arrange it in 10 factorial. And this was your previous answer, right? This was your previous result. So you'll say, subtract the previous result where your m of them are sitting together or m of checks are together. So when you subtract this result and this result, that means if you do n factorial minus m factorial times this, you will end up getting such cases where all the five will not be together. That means at the max four will be together. The one person will always be away. Yes or no? Yes or no? Any questions? Okay. Now, let me give you a question. Let me give you a question. Let's say this time there are six boys and there are four girls. Okay. In how many ways can you arrange six boys and four girls? In how many ways can six boys and four girls be arranged in a row, arranged in a row such that or so that no two girls sit together or no two girls are together? Question is clear. Question is you have to make these 10 people sit in a row such that there is no two girls sitting together. That means all girls have been separated out. Two ways. Karthik, somehow your answers are single-digit answer. Doesn't it raise the concern in your mind that they might be wrong? Two ways. I can count 10 like that only. What two ways? The problem is you are taking away the identities on these people. Every boy, every girl is a unique one here. So, what you are seeing is the placements. Who is coming in those placements will change the number of ways. Even if you take five girls, it will not be one, two way. How will it be two way? Tell me how will it be two way? Even if you take five girls, five boys, it will not be two way. It will be much more than that. Which girls starts won't it matter? What you are saying I understood your concern. You are thinking boy, girl, boy, girl, boy, girl starting with a boy or girl, boy, girl by starting with the girl. No. Which boy starts that is also going to make a difference. There are five boys. So, how will it be two way? It is not the gender of the person. It is gender as well as identity. Is it boy number one who is starting it? Boy number two starting, three starting and then these five position of the boys and these five position of the girls. They are all changeable. If they change the combination, the arrangement will be a different one. Yes. Any idea? Okay, okay Prisham. Okay Shah, will you take your time? We have time. Don't worry. Absolutely right Nikhil. Prisham has given an answer. Okay Prisham, can you justify it? Why did you get two factorial into nine factorial at the end? Is the question that I'm talking about, is it same as the this concept which I discussed right now or are they different? Is this concept and this question, are they different or are they same? That's what I asked. That's the reason I asked, are those two concepts same or different? If they're same, then your answer is probably right but I don't think so. They are same. They are different things. The reason why I gave this problem is because I wanted to see whether you are using the formula blindly because that formula doesn't correspond to the situation. Read the question carefully and read the previous concept carefully. Previous concept says those M specified things should not be together. That means M minus one can be together. M minus two can be together. M minus three can be together. But all of them should not be together. That is what this concept says. But what my question says is something else, question says no two should be together. They are not the same things. Yeah, unless until you put our very special value of M, which is two. Okay. So let's say you want to solve this question from a layman's perspective. How will you solve it? That is what we are basically trying to encourage everybody to adopt. So see here I will be using a method which I call as a gap method. So you already heard from me string method. Now gap method. Now see how does it work? So what do you want to separate here? Girls you want to separate? So first you make the boys arrange. So there are six boys B1, B2, B3, B4, B5, B6. Okay. How many ways just answer this question whatever I'm asking you. How many ways can you arrange the boys among themselves? What are the number of ways to arrange the boys among themselves? Just write it down on the chat box. Six factorial correct. Good. Okay. How many gaps are there between the boys? Count the number of X's and tell me how many gaps are there between the boys? Seven gaps. Correct. Now can I say if I place the girls in these gaps? Okay. And even I can arrange the girls among themselves doesn't matter. Can I say the girls will automatically get separated? So let's say girl number one comes here, girl number two comes here, girl number three comes here, girl number four comes here. So automatically you'll see that the girls have got separated. So the arrangement here would be like boy one sitting, then girl one sitting next to him, then boy two, boy three, then girl three, boy four, boy five, girl three, sorry this was girl two, girl three, the boy six, girl four like that. So if you place your girls in the gaps over here and of course you arrange those girls also, you would realize that you have achieved the purpose of the question that the girls are all separated out from each other. Now these gaps that I have shown, they are imaginary gaps. That means there is no chair there. That means wherever you put the gap, you automatically put a chair for the girl to sit there. Okay. Don't be thinking like there are already chairs in between them. No. They are imaginary gaps. That means there may not be a chair there but you're thinking as if there is a chair lying there. So let's say girl one wants to sit here, put a chair there for her to sit. That's it. Okay. So now think as if there are seven gaps, you have to pick four gaps from there and not only that, you have to arrange these gaps like arranging the girls. How many ways can you do that? So how many ways can you arrange four girls or four gaps? Let me call it gaps. Four gaps taken from seven gaps. Okay. Now these gaps are all distinct because in these gaps you're going to place distinct objects. So the gaps automatically become distinct. So that will be seven P four. So the answer to this question as per fundamental principle of counting would be six factorial into seven P four. I think the only one person who gave this answer correctly was Nikhil. Okay. Now there's another way to prove it, another way to solve it, but that would be quite lengthy, which is like exclusion case where you want to take total cases and you want to remove from the total cases those cases where any two girls is together, any three girls is together and all the four girls are together, but that would incorporate a bit of complexity while you are applying it. So this is the safest way to solve this question. So this becomes your answer. So remember gap method, very, very important and interesting method. Okay. So what I have exposed you is to some basic restrictive permutations and we'll take few more questions. I think one more we can manage to take in the next five minutes. Could you go up for one second? Okay. How is this, is this fine? Good enough? Great. So whatever we have done, we'll take some, I know questions based on the same and then we will call it as an array. Now in how many ways, in how many ways, in how many ways can these alphabets A, B, C, D, E, F, G? Okay. How many ways can these seven alphabets be arranged? Be arranged such that beg or CAD pattern does not appear in that arrangement, does not appear in that arrangement. Understood. What is the question saying? See, the question is saying you arrange it in, you know, you can arrange it in so many ways, right? But your arrangement should not be such that you can see a beg word in that or a CAD word in that. For example, let's say you arrange it like this. Okay. Let's say A, C, F, P, E, G, D. Now this is not allowed. Why? Because you can see beg word over here, correct? Okay. So just like the way you, you know, you know, sometimes Sudoku and all are there, you can find out a word there. Okay. So this should not be there. Okay. Similarly, they should not be a word like this. Let's say B, F, G, C, A, D, E. This also is not allowed because they can see a CAD word over here. Okay. So beg or CAD word should not appear in that pattern. And of course, obviously should not have a word like beg, CAD, let's say some B, G, F, A, C. What is left off? I think F is left off, right? This also should not be there because both beg and CAD word can be seen. So either this or this or both of them, not allowed. So how many such permutations of these seven letters can be done? So is that in no permutation, you can see a beg word or a CAD word or both? Of course, both and all I need not speak out. It is very evident. Just seriesing it because this is just to explain you and give your answer in a raw format. I don't want you to give me as a simplified number because then I cannot see the logic behind it. I'm interested in seeing the logic that you are using to solve it. Okay, Seethu, it's definitely wrong. Very good, Nikhil. Nikhil, you could have written that as five factor laws, right? Yeah, so four factor into five is five factor. Excellent, very good. Why is my Siri getting activated? Hey Siri, can you solve the question which I gave on the board? Siri is not responding. Okay, guys, let's discuss this. Okay, now first of all, let me ask you. Thirty seconds you want. Okay, I'll give you thirty seconds. See, your solving is more important, right? I am a facilitator. Ultimately, you have to achieve, right? Okay, Karthik, very good. Siddharth Karthik, Sharduli, excellent, excellent. No, I don't want the answer in that way. No, sorry. Give me the answer in a raw format. I want to see the logic behind it. Karthik, Sanoj, we'll discuss it. Okay, see, if I have to solve this question, what I will do first, I will find out the number of, okay, total words. Okay, you be the, let us say you be the total number of words. What are the total, I know, words you can make by using these alphabets only. Please note that you can make seven factorial words. Of course, I'm using all of them at a time and I cannot repeat them because there's only one A, one B, one C. So that's the meaning of the question. How many ways can you arrange these all alphabets? Means all of them have to be used and they cannot be repeated because they're all present one, one, one, one, you know, times in that expression. This is the total number of words. Now, let's say X be the number of or X be the set of words where big pattern appears. So let's say the words where big pattern appear. Okay, let's say Y be the number of words where CAD pattern appears. Okay, CAD pattern appears. Right. Now, what do I have to find out? Let me call this as universal set, number of elements in the universal set. Let me start a number of words. So I have to figure out in how many words the CAD pattern or the big pattern doesn't appear. In short, I am using my cardinal properties of sets. Now we all know that this is NX, NY minus NX intersection Y. Let's try to fill in this. This is only seven factorial, very fine. I mean, no worries about it. Now NX, NX means how many words are there where big pattern appears. Now let's take this word beg and tied with a string. Apart from this, we have I think AC DNF. Okay, so one, two, three, four, five, five units are there, which you can arrange in five factorial ways. Now here, please understand, you cannot or interarrange BeG because if you do interarrangement of BeG, your beg pattern will disappear. So no interarrangement of BeG. So this is the total number of ways in which a beg pattern will appear in a particular word. So that you need to subtract. Same will go for CAD also. If you tied CAD and make it as an entity, let's say I'll do that again here. So let's just CAD, I tied it up. And of course, other words are B, G, F, anything I'm missing, A, B, C, D, E I'm missing. Yeah. So again, five entities and you can arrange them in five factorial. Please do not interarrange CAD. Okay. Yes, sir. So now this is also five factorial. But remember, when you accounted for the words where beg pattern is appearing, or when you accounted for the number of words where CAD pattern is appearing in both of them, beg and CAD both of them can appear. So you have to remove them. That is what this formula says double counting would happen in this case, or double removal will happen in this case, isn't it? So I'll figure out how many of these words where beg and CAD pattern both appear. So I will take two strings now. Okay. And of course, there's one single alphabet F Oh, I'm so sorry. Yeah, one single alphabet F. So now these three words, the three units can be arranged within themselves in three factorial ways. So that it was an overcount in nx and ny so that you remove from here. In fact, you're blindly following the formula that you had learned in your sets. So this becomes your final answer and that will come out to be seven factorial minus two into five factorial plus six. Okay, of course, please you know, calculation is, you know, your domain I am not going to do it. But this is the logic. This is the expression that you should get. What is three P four, Karthik's knowledge, there is some, is this there's something like three P four. Can you choose four objects from three objects? Okay, anyways, thank you. I think one of you had a doubt. I think Tejaswini had sent me a doubt. So we'll take that up there just when you can stay back. Rest can leave if you don't have any doubt. Thank you so much. Good night. Take care.