 Today I will be talking about distribution of identical objects into different groups with upper and lower restriction. So up till now we were not talking about any such case where there is an upper restriction on the distribution of object. If you recall in the last session we had done few questions where we are trying to distribute certain identical objects. With lower restriction of course it was there but we did not put any upper restriction on the number of objects a person could get. So today I am going to discuss with those scenarios where there could be both type of restrictions in the distribution of object. So I would start with an example and then I will discuss the theory behind the concept. So let me begin with an example here. So let's take a question. Let's take this question. Now all of you please read this question very very carefully. This question says in an examination the maximum marks each of three papers is 50 and the maximum marks for the fourth paper is 100. So there are four papers altogether. Three of them have a maximum marks of 50. So as you can see three papers have a maximum marks of 50 and the fourth paper has a maximum marks of 100. The question says find the number of ways in which a candidate can score 60% of the aggregate marks. So everybody has read the question. Read it again. We will discuss it out in detail. Now when you look at this question you feel like solving this question by saying okay sir there are four papers. Let's say paper one, paper two, paper three, paper four. And let's say this person gets X amount of marks in paper one, Y amount of marks in paper two, Z amount of marks in paper three and W amount of marks in paper four. And I'm assuming that the marks that he is getting are integers and he cannot get negative marks in any paper. So I'm making two assumptions over here even though the question doesn't explicitly mention it. Number one assumption is there is no negative mark in the paper. Okay. Okay. And there is no fractional marks given. No fractional marks. For example, many a times we are given two and a half in some question, right. So something like that is not there. So a student can get only integer positive integer marks in every paper. Okay. So now you will say that okay sir this question is very simple. This is nothing but 20 60% of the aggregate. So what is 60% of the aggregates calculate that. So 60% of 50 plus 50 plus 50 150 150 plus 100 which is 250. So 50 60% of the aggregate will be nothing but 150 marks. So we'll have to solve this situation where X plus Y plus Z is equal to 150. Right. So if you think we can solve this question by using our previous formula in the previous class if you recall just go back to your notes. How do you distribute an identical objects into our different groups where there is a blank group allowed. So you do n plus R minus one CR minus one if I recall correctly. Yes or no. Can I apply that concept to solve this question. What do you think. What do you think. In the previous session we had learned a formula that if there are let's say our groups. Okay I'm just recalling your concept of the last class. And you're distributing an identical objects into our groups where your excise are greater than equal to zero means blank group is allowed. Then the total number of ways to do it was n plus R minus one CR minus one am I right. Everybody recalls this. Can I use this to solve this question. The answer to that is no we cannot use this concept to solve this question why. Because in this there is no upper restriction of course there is a lower restriction that it should be zero and above. Okay, if in fact if you talk about non-negative integers there is no restriction per se on this. Okay, so this method is not going to work out. This is not going to be helpful in solving this question because here there is a restriction and what is the restriction. The restriction is your XYZ is to be greater than equal to zero. But it should be lesser than equal to 50 that means there's an upper limit sitting over here. And the fourth paper should be greater than equal to zero and lesser than equal to 100. Okay, so this mechanism is not going to work. So it's going to fail in in such cases. Are you getting my point. So then how do I solve such a question to solve this question there is a very interesting I can say process which we call as a multinomial theorem that is going to be helpful in solving this kind of a problem. So let us talk about multinomial theorem and how will it be helpful in solving this problem. Okay, so guys and girls I would request you to be very very attentive. Because this concept has, you know, not been taken very well by students initially unless until you practice it out. Okay, so it is not a simple concept so as to say where you can let's say you know understand it by being unattentive also know it is very very you know I can say deep thought process so you have to listen to me very very carefully. Okay, so put down your pens. Okay, just lend me your ears and listen to me for the next five to 10 minutes of our discussion. Okay. Now, by the way the name multinomial theorem is just a misnomer, because actually multinomial theorem is much more, you know complicated we'll talk about it later on when we do binomial theorem. So we normally call it as multinomial theorem method. Okay, it is not exactly multinomial theorem it is the use of you know some kind of a term which resembles a multinomial expression. So what are these methods so see, consider that there is a term like this. See this person can get zero to 50 marks in paper number one right. So raise zero till 50 over some variability. Okay. In paper number two also, he can get anything between zero to 50. So raise a variability. Now I'll tell you why am I doing it most of you would be thinking are you why he's raising zero to 50 on you know some variability. Okay, I'll tell you the reason behind it. Just listen to me very very carefully. So since this person can score anywhere from zero to 50 marks. So in paper number one, I have created an expression which actually resembles a multinomial expression, having t to the power of those marks he can obtain. So t to the power zero because he could get zero marks t to the power one because he could get one mark t to the power two because you could get two more data data data data till t to the power 50 because he can get a maximum of 50 marks in the paper one. So I'll tell you the reason why I'm writing it like this. Okay, now let me explain this to you so pay attention. By the way, some of you are joining the classes very late. So I'll just copy pasting the same expression three times. Okay, but in paper four he can get from zero to 100 marks right so I'll be writing here. Okay, zero to 100. Now I'll tell you the reason why I'm writing it like this. I have no problem in you joining late but what will happen you will miss the initial understanding of the topic. And if you miss the initial understanding of the topic you will not be able to solve a question. Okay, it's like joining a movie in between, you will not understand what was the plot, why a person is behaving in this way, as it is shown okay. So please join it on time. I hope all of you are still having online classes right are you going, are you going for an offline class. How many of you are going for offline classes. Okay, so what time do you come back offline classes are running up till one o'clock two o'clock hundred o'clock. Oh, they just when he I can understand but yeah Kormangal is running till 1pm right. So 1pm to 4pm three hours is good enough time to come back have lunch and prepare yourself isn't it. Okay, so be on time please a sincere request. Yeah, now see everybody pay attention. When you multiply these terms, you will get, you will get all the powers of these starting from zero. Okay, I'm just putting some, you know, coefficient bracket over it, because some t to the power zero something into t to the power one will come so that something I'm putting in terms of unknown brackets. Okay, so when you multiply you'll end up getting you know something like this. Okay, so ultimately the last term will be t to the power 250 am I right. So t to the power zero all the way to the power 250 you will be getting by multiplying this but don't worry you don't have to sit and multiply this. Try to understand the motive over here. See you when you're multiplying systems what do we do, we pick up any one of the term from here. Okay so plus means or you either picked t to the power zero or you pick t to the power one or you pick t to the power two dada dada or you pick t to the power 50, that means one of the terms from this multinomial expression is picked up. One of the terms from here is picked up, one of the terms from here is picked up and then you multiply them isn't it. when you multiply terms where the base is the same, what happens to the power? What happens to the power? Power gets added up. Yes or no? Correct? Yes or no? So when you are obtaining let's say a term like t to the power 50, t to the power 150. So what are you doing? You are picking one term from here. Any term you can pick up, of course, it should be eligible to create. You can't put 0, 0, 0, 0 and create t to the power 150. Let's say you pick up 48 from here. Let's say you pick up 32 from here. t to the power 48, t to the power 32. And let's say you pick up how many? 80 is already taken care of. So let's say you pick up t to the power 30 from here. And let's say you pick up t to the power 40 from here. Can I say when I multiply these three, t to the power 150 will get generated. Isn't it? So when you multiply this, what happens? The powers will get added up. So 48 plus 32 plus 30 plus 40. So it will give you t to the power 150, isn't it? So that is one way to create t to the power 150. That is not the only way to create. Okay. There are so many ways by which you can create t to the power 150. Right? Yes or no? Let's say you pick up 50 from here, 50 from here, 50 from here and 0 from the last one. That will also give you t to the power 150. Let's say you take 40 from here, 40 from here, 40 from here and 30 from the last one. That will also give you t to the power 150. Now, understand here, this bracket which I have left, this bracket will basically give you the coefficient. Okay. So this will give you the coefficient of t to the power 150. So some number will come there. This number actually gives you the count of the number of t to the power 150 is created in the product. Yes or no? So what is the coefficient doing in front of t to the power 150? So coefficient is like a counter. It keeps a count. It keeps a count that how many t to the power 150 terms are created. Isn't it? Isn't it? See, when I say one plus t to the power two, what do we write it? One plus t plus t plus t squared. So when you multiply one plus t with one plus t, let me write it like this. Okay. So this number two that I have got over here, this number two keeps a count of how many t's are created. So they're two t's created. That is why there is a two t coming up. There's only one t squared created. So the coefficient is a counter. He's a person who is sitting and keeping a count. Okay. So whenever t to the power 150 is created, he'll do plus plus. Like what do you do in your computer coding? Isn't it? So this guy, the coefficient is basically the count, keeps a count of how many t to the power 150 will be created. Basically, this is the number of ways in which one can obtain 150 marks in the paper, in the four papers. Yes or no? Do you all agree with me or not? So if I get this coefficient, my job is done. That coefficient is my answer straight away. First of all, everybody agrees with me on that or not. This itself is very important. Understanding this itself is very important. So when you multiply these terms, you are indirectly taking the marks of this guy from these four papers and you're adding it and you're trying to see in how many ways he's getting 150 marks, which is that 60% of the aggregate, right? He's supposed to score 60% of the total, right? Which happens to be 150. So that coefficient of this term is basically keeping a count of how many ways he's getting that 150 marks. Exactly. So indirectly, Siddharth, that coefficient, which is keeping a count of how many ways t to the power 150 is created is helping us to solve the problem because this 150 is nothing but the number of ways he can get that 150 marks. Are you able to understand? That is why I said even the sharpest of brains find it difficult in the beginning. Listen to me once again. Listen to me once again. Okay, let's say I get the coefficient of t to the power zero as one. What does this one indicate? Siddharth Karthik. This one indicates that there is one way in which you can get t to the power zero. And what is that one way? If you multiply t to the power zero with t to the power zero with t to the power zero with t to the power zero, right? Is that the only way? Correct? That means there is one way in which this person can score a zero in all the zero in the aggregate. If he gets a zero in the first paper, zero in the second paper, zero in the third paper, zero in the fourth paper, try to relate it to the marks now. Are you getting my point? So this power that you see is the aggregate marks that he is going to get. Okay. And this is going to keep a track of how many ways he can get, number of ways he can get this aggregate marks. Are you listen to me? Then only you'll understand. No, you are, you have given, you have assumed that you have not understood. Try. So chose. Think. Okay. I'll ask you a simple question. Tell me in how many ways can you create t to the power 150? In how many ways can you create t to the power 150? You'll say, sir, so many ways you already give two, three examples, 48, 32, 30, 40, that is one way of creating it. Correct. Now that one way will make it, will make this count plus one. Okay. So this will, let's say one. Okay. Let's say there another way where you have taken 50, 50, 50, zero. That's one more way. Correct. Another way would be let's say 20, 30, 40 and how much? 20, 30, 40 and let's say 60. That's another way. So what is happening? You start getting some numbers over here. That number is the coefficient of t to the power 150. So the number of ways the person is getting 150 in total will be same as the coefficient of t to the power 150 in this expansion. Are you now getting it? So I am relating the marks counting to the coefficient technique. So I'm just connecting those two techniques. Okay. These are just terms which let's say if I ask you how many ways can he get two marks in the paper? Tell me. Who will tell you or which term here will tell you how many ways can he get two marks in the paper? This term will tell you how many ways he will get two marks in the paper because this too, the number of ways you can generate an aggregate of t to the power two will tell you in how many ways you can get a sum of two or aggregate of two marks in all the four papers. So one way he could get is one, one, zero, zero. Another way he could get is one, zero, one, zero. Another way he can get is one, zero, zero, one. Another way he can get is zero, zero, one, one. So whatever are these numbers coming up? I mean, let's say they are 10 in number. This coefficient will actually become 10 then. I'm just giving an example. Actually, what is going to come? That is something which I will have to figure out by solving it. Is that part clear to you? In fact, is that part clear to everybody? So coefficient here, I'm again repeating it. Proficient here is basically keeping account of how many such terms are generated indirectly keeping account of how many ways can one get an aggregate of two marks in those four papers or aggregate of whatever terms you're willing to find it out. Yes or no? Okay, let me give a simpler example. Let us say there is a, there are three papers. Okay, I'll just give you a simple example. Each paper has a minimum of zero and a maximum of two marks. Okay, this also minimum of zero maximum of two marks minimum of zero maximum of two months. Okay, I'm just taking a simple example. Okay, tell me in how many ways can a person score in how many ways can a person score? Let's say three marks. Okay, so let's say if you virtually count it, you tell me how many ways can you score three marks? Let's say paper one he gets two, paper two he gets one. He gets a zero here. That's one way. Correct? Yes or no? Then two, zero, one again, one way. Then let's say one, one, two, zero, another way, one, zero, two, that's another way. Yes or no? Okay, zero, one, two, zero, two, one, one, one, one. Is there any other way to get a, is there any other way to get a sum of three which I missed out? You can highlight it out. Anything that I missed out, you can highlight it out. Is there any other way he can get a sum of three marks in this paper? No, right? Okay. Now this is what I'm trying to solve by taking a situation like this. So I'm creating such terms here. Okay. And I'm trying to multiply these all. When you multiply these all, what is the lowest power of T that you will get? Zero. It will have some coefficient. Okay. Then you'll have one, it will have some coefficient. Two, it will have some coefficient. Three, it will have some coefficient. And finally, you'll have some coefficient for four. Correct? Some coefficient for five, some coefficient for six, that is the max you can get. Okay. Now, if you actually multiply it. Okay, let me show you the multiplication also. If you actually take a cube of it. Okay. You would realize that the coefficient of T to the power three will actually come out to be this count which happens to be seven. This term will come out to be seven. If I have not missed out anything here and let me show that on GeoGebra. GeoGebra actually helps you to count slightly faster. Okay. So let me just expand that term. Yeah. So let me write f of x small caps f of x equal to one plus x plus x square to the power of three. Okay. Expand expand f of x. Okay. Now do you see that when I expanded it, you get the coefficient of x cube by seven. Do you see this? Please notice this. Do you see that? So indirectly, it is trying to say that there are seven ways in which T to the power three will be created, which means if you start adding, if you start seeing the aggregate of these three papers, see what are the powers? Powers are the marks, right? So when you're multiplying it, marks are already getting added up. You only told know if the base is same, powers will get added up. So it says that T to the power three, that means an aggregate of three will be obtained seven times. That means in seven ways, one can get T to the power three term and T to the power three term will come only when you pick these combinations over here. Are you getting my point? That means if I take T to the power two from here, T to the power one from here and T to the power zero from here, then only I will get T to the power three. That is one way to do it, which is basically hinting towards this marks that the student has obtained. Another way to do it if you pick T to the power two from here, T to the power zero from here, T to the power one from here. That's another way of getting T to the power three, which is basically hinting towards this marks which the student can obtain. Are you getting my point? So the number of ways in which you can generate T to the power three, there are seven such ways to do it. That seven basically tells you how many ways he can obtain three marks aggregate in these papers. So this simpler example, does it make it clear to understand the previous page example? No, it is clear. No, it is not clear. What doubt you have? That is a different thing, Sharduli. I'll come to that. I will tell you. Don't worry. First, you get the naksha of your house ready. Then how do you build that structure? Is a civil work? We will decide that. Okay, so this is understood. Okay, let's go back to the problem. Now in again, I'll rewrite the same expression. So in order to solve this, in order to solve this question, okay, in order to solve this question, in order to solve this question, I'll rewrite the expression once again. With special attention to Tejaswini. Oh, sorry. So Tejaswini, we had discussed that we need, what is the coefficient of T to the power 150 in this? You're right in saying that we don't need T to the power zero, T to the power one. Why am I writing it? I was just writing it to explain you the concept. Right? What I actually need from this product is what is the coefficient of T to the power 150 only. Rest I don't need. Maybe you are like confused device that is writing T to the power zero, T to the power one, T to the power two, right? That was just to explain you. But actually, what do we need? We just need the coefficient of T to the power 150. Okay, so out of all the terms that you will get, I just need what is the coefficient of T to the power 150. This is what we need. Okay. Other terms, I don't need. Okay, so I just have to focus on getting T to the power 150. Correct? So this is something where everybody is on the same page. Right? Say yes. If you have understood till this step. Okay. Now, do I literally multiply to get that T to the power 150? Definitely not. I would be a, you know, I would not be a smart person to do that if I am getting it in that way. Okay, so what I'm going to do, all of you, please, you know, pay attention. First of all, I'm going to bring your attention to the fact that you have an expression like this. Many people ask me, say, why did you choose T? I choose any variable you want. T, P, X, Y, Z, W doesn't matter. You just have to keep the name of the variable same. It doesn't matter what name you choose because you just have to get the powers added. Okay. So in this, we are finding the coefficient of T to the power 150 in this. Now, if you see this clearly, this is a geometric progression. This is also a geometric progression. What is the sum of terms of a geometric progression? We already know that A, correct? 1 minus common ratio to the power of number of terms, which in this case is 51 divided by 1 minus common ratio, correct? So this whole thing cube, okay? Similarly, this will be A, 1 minus T to the power number of terms, which is 101 divided by 1 minus T. Okay. So what are we doing? We are going to find coefficient of T to the power 150 in this. Clear? So far, so good. Keep me updated. If you're, you know, lost track somewhere, immediately bring it to my notices that this is where I lost track. Can you repeat once again? So so far, no issues at all. Okay. Now let me expand this. So I'm finding coefficient of T to the power 150 in 1 minus T to the power 51 cube, okay? divided by 1 minus T cube. This term will be 1 minus T to the power 101. 1 minus T, okay? In short, if you see you have got a binomial expression here. So that's why I was telling that multinomial name that is given to this method is a misnomer. Misnomer means it is not going to be exactly multinomial theorem. Okay, now all of you please pay attention. This is again another bottleneck that we will be getting. Any problem, any problem so far? Any problem somewhere? Tell me where? Second step, this one. Some of a GP, what is some of a GP? So up. Some of the GP is what? A 1 minus R to the power n by 1 minus 1. Who is a here? T to the power 0. 1 minus. What is R here? T total number of terms is 51 by 1 minus R. So where is the problem? Got it now? No issues. Anybody has any issues still there? Okay. Now, all of you please pay attention. Here also some amount of listening is required for me. Listen to me properly. See, if you expand this term, the first term, it will be expanded like this 1 minus 3 T to the power 51 plus 3 T to the power 102 minus T to the power 153. Am I right? The first term only I'm expanding it. Anybody has a doubt in that? Other terms I'm just copying as it is. Meanwhile, I have to just copy this. I'm finding this in this. Anybody has any issue in this? Okay. Now, let us multiply these two terms. Now, keep in mind that I need coefficient of T to the power 150. This should be running in your mind. So when you multiply it, let's see what do we get? So one multiplies to these two, you will get 1 minus T to the power 101 minus 3 to the power 51 multiplies with these two. You will get minus 3 to the power 51 plus now see here plus 3 to the power 3 into T to the power 152. Do you think this term is going to contribute towards T to the power 150? Do we need this term? I mean a power which has already surpassed 150. Will it be required for us? Will it be required for us? No, right? There's just any other questions. So how did you get T to the power 102? I have not written T to the power 102 so far. Oh, how did I get this? I have to teach you people how to write a plus b whole cube also. Chalo, I will do that also. This is a cube minus 3 a square b. Okay, then plus 3 a b square. Okay, then minus b cube. What is this they just meaning? Eyes and girls. If you ask such questions now it breaks the, I mean, okay, anyways. Yes, so coming back to this, is this term required? No, it is not required. So don't write it. So something which is not going to generate T to the power 150 simply because it has already surpassed 150. Why do we even write it? Okay, see ultimately we have to generate T to the power 150 is from the product. So if something has already surpassed T to the power 150 need not write it. Okay, so I think this end to this is multiplied now. This into this will give you 3 into T to the power 102. Okay, this into this will give you minus 3 to the power 203, which has already surpassed, which has already surpassed 150. Similarly, T to the power 150, this itself has surpassed 150. So we need not multiply. So there is no other term other than these four, which are of interest to me. Right. So from these terms, I have to look for coefficient of T to the power 150. Yes, or no, any doubt so far. Now don't ask me doubts like how did you get A minus B whole cube and all those things that doubt you first ask yourself how did how did I get this? Wait a second. We'll come back. Welcome to all those terms. I can understand your doubts. Okay, let's know this will not bring down the power ever. I'll tell you why. Sharduli where you want me to scroll up? This is the beginning of the page. Okay. Okay, done. Fine. So I'll do one thing. I'll just copy paste this in our next slide. I need to drag. So I'm just copy pasting the same stuff back. Now this term that you see here. Now try to go back to your bridge course days. If you have a term like this, just try to recall. Okay, recall during the bridge course. If you have a term like this, where this is a negative integer, right? So what kind of a binomial term it'll give you? It'll give you an infinitely long binomial expression, right? Let me write it down for you. It'll give you one plus one plus minus n into minus t. Then minus n minus n minus one minus t square by two factorial minus n minus n minus one minus n minus two into minus t cube by three factorial. And will this ever end? No, this will go on forever. Yes or no? So if you recall, I had discussed this with you that when there is a binomial term where the power is a negative integer or a fraction, that binomial expression doesn't stop anywhere. So the reason why I did not expand it in the first go is because this is a infinitely big binomial expansion. This is an infinitely big or infinitely. It has got infinite number of terms. Let me write it like that. Infinite terms are there in this. Yes or no? Do you all recall that fact which I've done in the bridge course? If you were not there in the bridge course, now please understand here that such an expression is going to learn infinitely big. Why does infinitely big? Why doesn't it stop like one minus something to the power of four? Because the term stopped because you started getting a zero in this particular coefficient. In this case, you will never get a zero. It'll keep reducing, reducing, reducing like that. In short, what you have written is an expression of this nature. And as you can see, these terms are not going to die out. These terms are not going to die out because this n is a number which is positive number. Negative n is a negative number of course, but n is a positive number. It's not going to die out. So it's going to run up till infinity. Are you getting my point? Yes or no? Now, all of you again, pay attention here. If I ask you this simple question, what is the coefficient of t to the power, let's say, k in this expansion? What will your answer be? Any k? k could be 1, k could be 0, k could be 3. What will the answer be? So you'll say, sir, it will be n. Let's say my k is 3. So it'll go all the way till all the way till n plus k minus 1 by k factorial. Am I right? Yes or no? Negative sign got cancelled out, say to here. See negative n negative t became plus nt, right? Here also you have two negatives and one negative here is squared up. So that is also gone. Here you have one negative, one negative, one negative and three negatives from here. That is also gone. Yes or no? Correct? Okay. So can I say, looking at this binomial expansion, the coefficient of any t to the power k term in this expansion will be this? Yes or no? Now, let's do one small activity. Let's multiply with n factorial in the numerator and denominator. So I'm just writing the same term back and I'm multiplying it with n minus 1 factorial. Correct? So what does the numerator become? Can somebody tell me when you're multiplying this on the numerator? What does the numerator become? What happens to the numerator? It becomes n plus k minus 1 factorial. Correct? Divided by n minus 1 factorial k factorial. Can I write it in shortcut as n plus k minus 1 ck? Now, remember this result because it is going to be very useful for us in solving the given problem. Okay. So first, note this down and if you have any questions so far, do let me know. Then I will go to the original question. Still the question is not solved. We are still working on the where with halls of this problem. So I still have to solve this question. So but notice, sorry, note this down first. This doesn't look good. So I'll just, okay, note it down. Okay, now let's go back to this question. In fact, I will again go to the next slide. No, let me not go to the next slide. I'll just manage over here. Okay. So let's say I have this term and I'm copy pasting it again. So from this infinitely long series. So there are these terms. Okay. And this is an infinitely long series. This will have so many terms. Okay. Now do I have to write all the terms? No, I will have to categorically pick up terms which I actually need to create t to the power 150. So see here, if there is a one sitting here, which term do you require from this infinitely many terms to generate t to the power 150? So you will say, sir, you need something like t to the power 150 from here. Am I right? Yes or no. So out of these infinitely many terms, this is one of the terms which I need. Right. Now you must be wondering, sir, why, why haven't you written any, you know, value in front of t to the power 150? Then that is something which I want you to tell me. In this expansion, what is the coefficient of t to the power 150? The answer is in this formula, which you have already derived. So n is for case 150. So can I say the coefficient there will be 153 C 150. Am I right? Yes or no. So this term that you require here, that will be having a coefficient of now let me write it down 153 C 150. Yes or no. Have you understood? First of all, why do I require t to the power 150 from here and what will be the coefficient of hair that also is clear to you? Anybody has any doubt over here? See, this is an ocean. 1 minus t to the power minus 4 is a big ocean. So many terms are there. Infinitely many terms. Do I need everything? Do I need everything from the ocean? No. What do I need depending upon what do I have in these terms? So I have a one here. So I need t to the power 150 related term. Then only these two will multiply to generate t to the power 150. Isn't it? Yes or no. So they are like now supplementing each other or complimenting each other. They're trying to create t to the power 150 together. Yes or no. Okay. Now tell me what other term will I need? I have t to the power 101 sitting over here. What do I need from here? Dot dot dot dot dot. What term do I need from here? Something t to the power something t to the power 49 absolutely nickel. But what is this something again for that this formula will help us. So n is for K is 49 C 49. So this is going to be 52 C 49. Am I right? So this coefficient will become 52 C 49. Let me just remove these brackets because we don't need it. Is it fine? I'm putting dot dot dot dot dot means there are so many other terms. I don't care about them. I only need these ones. Okay. Now again, look at this term and tell me what term do you require from that infinitely many terms dot dot dot. What term you require t to the power? Tell me t to the power 51 is here. So in order to complete 150, what power do you need 99? Isn't it? So you need something t to the power 99. Now who will tell me what is this something that I have left over here? Fill in the blanks. Use the same formula. Now n is for K is 99. Tell me what will come in n plus K minus one C K. What will come? Right? So 102 C 99 will come. Is it fine? Any questions? Yes or no? Okay. Now this time I don't want Nikhil to reply. I want somebody else to reply. I already have t to the power 102. So in order to complete t to the power 150, which term is desired from that ocean? Something containing which power of t other than Nikhil? Somebody answer this. Right? Shardari t to the power 48 and what is the coefficient of that term? What will be the coefficient of that term? n is for case 48. Tell me what is that coefficient? I put it in the formula. No 4 plus 48 minus 1 C 48, 51 C 48. Absolutely Siddharth. Okay. Dot dot dot dot dot dot dot dot dot dot dot. So beyond this I don't need anything. Right? So again I'm repeating this term is like an ocean. Ocean full of let's say you know fishes do I need every fish know I need the ones which I can eat. Isn't it? In an ocean, there are so many fishes. Okay, so many water, water animals. Do we eat all the fishes? No, maybe the ones which we, you know, suits our taste. Okay, so I'm only catching those fishes, which will help me, which will help, you know, to quench my hunger. Isn't it? So now, if you want to generate T to the power 150, this guy should multiply with this guy. So one should multiply with 153 C 150 into T to the power 150. So this is one coefficient. Okay, this term should multiply with this term. So that will give you minus 152 C 49 T to the power 150. This term should multiply with this term, which will give you minus 3 into 102 C 99 T to the power 150. And this term should multiply with this term, which will give you 3 into 51 C 49 T to the power 150. There really so many terms generated, but I don't need them. I only need T to the power 150. So overall, what is the coefficient of T to the power 150? This is your coefficient. So what does it actually come out to be? So one into this, which is 153. By the way, 153 C 150 is as good as 153 C 3. Then you have minus one. No point writing minus one, just like minus 52 C three. This will be minus 302 C three. And finally, three into 51 C three. Okay, this is 48. Yes, I'm so sorry. Is it fine? Any questions, any concerns anybody has? Now, it is up to you to evaluate this. Okay, and this can be evaluated. You just have to slightly be patient in calculating it. But if you just see the coefficient, I'll just tell you the answer that comes. It comes out to be 110551. You can check it out with your calculation devices also. Okay, so these many ways that guy could have actually scored, could have actually scored 150 marks. Is it fine? So see one question actually took us 55 minutes, because of course, we understood the entire procedure. So that is why this is one of the most trickiest concepts in permutation combination. Okay, so you need to practice more of such type. There are of course, questions in DPPs, which I'll be sharing you after the class. Okay, so it requires a bit of practice. I remember last last year, we had a very, very smart kid. I think his rank in JEE was some 1900 rank. He actually did not understand this concept till it was again revisited in the crash course. Okay, because he could not, he did not practice it. So you have to practice this out. It requires a bit of practicing. Okay, so I will not be taking any further problems on this. Okay, but if you have any questions, please feel free to ask me. Who told you is lesser than one? From where you got that idea that it is lesser than one? No, I didn't get you why you got an idea that it has to be lesser than one. Do you know the value of TA? So why and see, are we worried about the expression value or are we worried about the number of terms in there? Even if you have infinite amount of terms, the sum can be finite. So how it has something to do with the powers that you're getting. There's no relation. Nikhil, you're trying to compare mangoes with oranges. That's a different thing where you're trying to talk about the value of that expression. That is none of our concern right now. Our concern is only that this is a binomial term which has got infinitely many terms. Whether those infinitely many terms add up to give you a number less than one greater than how does it matter in this question, right? That's a different agenda altogether. Maybe we'll talk about it in our binomial theorem chapter whenever we do it. Can you find a number of ways to sum up a number? I didn't understand that question. What is the meaning of number of ways to sum up a number? Sum up a series you should say, find the number of ways to add 50s, add up to 56. What do you want to add up to 56? See, you're not getting the basis of this formula. This basis of this formula, I'll again take you to the first sheet. The basis of this formula is there are certain things which are adding up to certain value, but there's an upper and lower limit to it. When you're adding natural numbers, is a natural number having an upper and lower limit? Two is two, my dear. Okay, here X and Ys are having some upper lower values, like X can be between these values, let's say somewhere between 10 to 55. Of course, taking integers in between. Y can be somewhere between, let's say 17 to 57. Z can be somewhere between 1 to 42. I'm just giving some dummy example. So how many ways can you select the values which are given by these ranges to generate a sum of 150? How does it fit in into your requirement of adding natural numbers? That is not clear to me. Maybe if you can frame the question in a better way, I will be able to help you out. So last, not the last, the second last topic that we're going to talk in this chapter is geometrical problems. Okay, geometrical problems. And after this, I'm going to talk about derangements. I think some of you wanted to, some of you had a doubt in derangement questions. We will talk about that as our last topic. Now, geometrical problems are the favorite of schools as well as competitive exams. So there is nothing separate to be discussed here. We just have to solve few questions. Okay, so those questions will probably try to understand the concept. So let's have a question. Let's take this question. I'm sure you would have seen this question in your school papers also. There are 10 points in a plane out of these points, no three are in the same straight line except for right. So out of these 10 points, there are only four such points which are collinear apart from those four, no three are collinear. Okay, first part of the question is how many straight lines can be formed? So basically, it's like something like this. So I'm making some random 10 points here, one, two, three, and I'm making some four of them collinear, one, two, three, four. Okay, one, two, three, four, five, six, seven, eight, and let's say nine here, 10 here. Okay. Now, no three are collinear other than these four. That means these four are in the same line. But apart from it, there is no such three points from where you can actually draw a straight line. Then tell me, how many straight lines can you make joining these points? That is the first part of the question. Okay. So now, how many points do you need to create a line? What are the minimum number of points you need to pass a line through them? Yes, I know. How many points are required to? Yeah, absolutely right Siddharth. So out of 10, you choose two points through which you can pass a line, correct? But if you recall, four of them are collinear. So these four, when you say four C2, this is not going to give you six such lines, this is only going to give you one line. So what do we do? We subtract this and add a one. Are you getting a point? So this answer will be 45 minus six plus one, which is 40 lines can be created. Is it clear? Any questions, anybody? Any question, any concerns? For the first part? Okay. Second part of the question, how many triangles can be formed? See, in 10 C2, you are assuming that no three points are collinear. But in reality, is that the case? No, right? Four of them are collinear. So in this 10 C2, this will be some extra cases. This will be, you can say over counting that you are actually doing because these four points will not generate four C2 points. These four points will only generate one line. So what do you did? You subtracted these because these are extra counting that you have done. But when you're subtracting this, by mistake, you're also subtracting that one line that they can actually have in between them. So see what is happening? You're considering that this is a separate line. This is a separate line. This is a separate line. This is a separate line. But this is not separate, separate, separate, separate line. They're not six different lines. They're actually one line only. Right? So this was an over count that occurred in this. Is my jada count? Are you getting my points? So in 10 C2, there was surplus. We should not have been there. Yeah, this, this, this basically considers that six lines to be different lines. But in reality, they were not different. They were only one line. Hence you subtracted them. See if, if four points are collinear, you should join these two and make a line. Line is an infinitely extending object. So you choose any two lines out of these four, two, sorry, choose only any two points out of these four. Are they going to give you four C2 lines, six lines? No. They're only going to give you one line. Yes. How many triangles can be formed? For triangle, the same concept, you will say take any three. But these four will not, these four, which I had shown here, again, I'll show it once again. Sorry, I put these four will not give you four C3 triangles. So you have to subtract that. Okay. Yes, absolutely. This will be 120 minus 416. Correct. Sharduli. Let's do the fourth, sorry, third one. How many quadrilaterals can be formed joining them? Okay. Sharduli. Okay. All of you please listen to this carefully. That is not correct, by the way. Siddharth also, that is not correct. So all of you please listen to this. Okay. If I take 10 C4, it will give me, it will give me basically the number of ways in which I can select four points out of 10 to create a quadrilateral. But try to understand when a quadrilateral will not get created. A quadrilateral will not get created if you take all the four from here. Correct. No. So can I say I have to subtract that case? Correct. Or three from here and one from somewhere else. So can I say this is a quadrilateral? You'll say no, sir, it is a triangle. So you take three from these four and one from the remaining six that will not create a quadrilateral for you. That will actually create a triangle. Are you getting my point? So if you take any four from here, that is four C4, that will not create a quadrilateral. So you need to subtract that. If you take any three from these four and one from the remaining six altogether, 10 points are there. No. Four is here. Six is outside. I mean six are other scattered somewhere. So if out of that six, if you take one and three from here, any three, it is not going to create a quadrilateral. Correct. Yes or no. So these are the cases which you need to remove. So six C4 is 210 minus one. And if I'm not mistaken, this is 24. So this answer is 210 minus 25. They can be only 185 quadrilaterals formed. Is this fine? Any questions? Any concerns? Got it? Everybody? So these type of questions are also commonly asked. The next type of question that we are going to talk about is counting of rectangles and squares in a grid. So we'll take that up also. Any questions anybody? Okay. So grid questions. Grid problems, I call it. So let me give you a grid here. I'm just, you know, making a orbit number. Let me change the color so that I can refer to those lines by the colors. So as you can see here, I have made a, now focus on this part. Focus on this part. You can see here that there is a grid. Okay. Just focus on this part. Don't focus on the parts which have got extended from there. Okay. So this is a, I can say four by five grid. Okay. Four into five grid structure. Okay. Four because it's along the x axis, four along the y axis, five. Okay. My question is the following number one. How many, how many rectangles are there in this grid? Okay. How many rectangles are there in this grid? How many squares are there in the grid? Okay. I will give you further more questions, but let us try to solve these two first. By the way, rectangles means including square. Rectangles includes square as well because all squares are rectangles. I hope everybody knows that. All squares are rectangles. Okay. So tell me how many rectangles are there in this grid? Of course, one way is to literally count them, but I am looking for a way to solve this problem rather than solving this particular problem itself. That means if I extend it, let's say, if I make a big grid, big mesh, how many rectangles will be there in that mesh? Yes. Yes. Why not? This whole blue thing is a rectangle only. Any idea how to count rectangles here? No. Okay. Now listen to this everybody. If you see, I have created this grid by using some white sticks. Okay. These are your white sticks. I'll call them as vertical sticks and using some horizontal sticks. Isn't it? The yellow ones. Okay. Now in order to create a rectangle, can I say I have to pick any two of the white sticks and any two of the yellow sticks? So for example, assume that this structure is laid over bamboo sticks. So there are five bamboo sticks which are white and there are six bamboo sticks which are yellow in color. So let's say if I pick up this bamboo stick and this bamboo stick from the white ones, and let's say if I pick up this bamboo stick and this bamboo stick from the yellow ones, I will be getting a rectangle trapped here. Am I right? So let's say somebody comes here and he lifts these two white sticks and somebody comes here and he lifts these two white sticks. Can I say a rectangle will be trapped in between those white ones and the yellow ones which these guys are picked up? Okay. So can I say the number of rectangles that you can basically form is how many two sticks you can pick from the whites and how many twos you can pick from the yellows. So from the whites you can pick five C2 sticks because you have to pick two out of five and from the yellow ones you can pick two in six C2. So altogether your number of rectangles that will be there in this grid will be 10 into 15, 150 rectangles will be there. Are you getting my point here? Is it clear? Is it clear? Okay. Now let us scale this out. Let us scale this up. I always prefer you learning by taking a simpler case and then scaling it up. So let us say if I scale this up. That means in general, in general, if there is a grid which is formed by, let's say let me give you a, let me give you a, answer this question very, very carefully. Okay. Don't rush into solving it. Okay. So let's say this is your white lines. So let's say this is a m cross n grid structure. That means this length is m and this length is m. Okay. Tell me how many rectangles are there in this grid? What will your answer be? What will your answer be? Very good. m plus 1 C2 into n plus 1 C2. Please note many people forget this plus 1 plus 1. Please remember to make a grid structure of m by n, you need m plus 1 vertical sticks and to make, you know, this thing you need n plus 1 horizontal sticks. So don't forget this plus 1 as your result will go wrong. Okay. Now coming back to our, how many squares are there in this? How many squares are there in this 4 by 5 grid? Please don't say 20, right? Because 20 is something which is like 1 by 1. 1 by 1 size, 20 squares are there. Agreed. 2 by 2, 3 by 3, 4 by 4 also are possible. Right? Okay. So think carefully and tell me how many squares are there in this 4 by 5 grid and how do you count that? Anybody has any idea? You can discuss it with me. Okay. Okay. So Nikhil, let's try to answer this question by breaking this event into cases. Case one, how many 1 by 1 squares are there? How many 1 by 1 squares are there? Now, of course, you'll say, sir, 4 into 5, 20. Okay. But let's have a technical way of counting it. A 1 by 1 square will be created if you pick any 2 adjacent horizontal and vertical sticks, isn't it? So a 1 by 1 square will be formed only when you pick up any 2 adjacent vertical sticks. Like, means you can pick these 2 or these 2 or these 2 or these 2. That means there are 4 ways to do it. 1, 2, 3, 4 or you pick any adjacent horizontal sticks. That also 1, 2, 3, 4, 5. So 1 by 1 square will be when you pick up, when you pick up any 1 of the 4 on the top and any 1 of the 5 in the horizontal position. Yes or no? So that is why you ended up getting 20 as your answer. Okay. Tell me how many 2 by 2 squares will be created? How many 2 by 2 squares will be created? Tell me. So for 2 by 2 squares, all of you please pay attention. I'm just erasing this off. You have to pick up any 2 vertical sticks which are at a distance of 2 units apart. That means you can pick these 2, these 2, these 2. That means 3 ways to pick up vertical stick and you have to pick up any 2 horizontal sticks which are 2 units apart. These 2 or these 2 or these 2 or these 2. That means 4 such cases. Okay. So you will end up getting 12 2 by 2 squares. 2 by 2 squares means dimension of 2 by 2. Everybody is it clear to you? Does it make sense? Makes sense, everybody? Yes, no, maybe. Okay. You tell me how many 3 by 3 squares will be generated if you have understood it. So what is the number of 3 by 3 squares? That means 3, 3. Yes, tell me. Right, Nikhil. 2 into 3. Correct. So for that, what do you do? Simple, same technique. Technique is not going to change. So you have to pick up such 2 sticks which are at a distance of 3 units apart. So 1, 2, 3, 3 units apart. So these 2 can be picked or these 2 can be picked. Here also you have to pick up such sticks which are 3 units apart. So these 2, these 2 or these 2. So 2 from here, 3 from here, so 2 into 3. How many 4 by 4 squares you can pick up? How many 4 by 4 squares you can make? That will be similarly 1 into 2. Can I make a 5 by 5 square? You'll say no sir, one of the dimensions is only 4. You cannot go to 5 by 5. So your answer together will be just the sum of these which is nothing but 20 plus 12, 32, 32 plus 6, 38, 40 such squares will be there. Is it fine? So see 150 rectangles out of which 40 squares. So how many non-square rectangles will be there? That also can be asked to you. So 110 non-square rectangles are there. Is it clear? Any question in this approach? Please ask me. Okay, that is very important. 100 by 100 grid, there is a process. 1000 by 1000 grid will tell you. Will tell you, don't worry. Can you also take a case and I pick blue line 1 and 3 and I have to pick yellow line also in the same manner that gap between them is 2. By the way, this blue structure is a overlap on this. It's actually white line. If you want to form a 2 by 2 square, that's what you're saying Sharguli. If you take 1, 3 from the vertical and 2, 4 from the vertical, from the vertical only you're saying. So didn't I take that case and that 2 by 2 squares? Really didn't get you. I have already accounted for that in this 3 into 4. Generalize case. Let's talk about generalized case. So in such cases, we first have to understand, we first have to understand which is smaller of the two. For example, here you can see that this was a 4 by 5 grid. So you need to understand which is smaller of the two in this case. So let's say m is smaller than n. So the total number of squares that will be formed will be m minus times n minus r, r going from 0 to m minus 1 provided m is smaller than n. Is it correct? Can I say this is going to be your answer? Any questions? Any questions? Any concerns? So please note this down. And after you have noted this down, I'm going to ask you a question. Done everybody? Okay, ready for a question? Okay, so the question is there is a chessboard. Okay, I hope everybody has seen a chessboard. Okay, so a chessboard has got. Okay, some of you are players of chess. You would have seen chessboard. So let's say this is a chessboard. Chessboard is actually a 8 by 8 grid. Okay, so I have the following questions for you. Number one, how many rectangles does the chessboard have? Number two, how many squares does the chessboard have? Okay, number three, how many rectangles, how many rectangles or how many non-square rectangles are there in the chessboard? How many non-square rectangles are there in the chessboard? And number four, okay, I'll come to number four question a little later on. As of now, I will not ask you. So please give me these responses within one and a half minutes because this is all done. So first one, first, how many rectangles are there on a chessboard? Wrong, chargouli, wrong. Right. This value comes out to be 9C2 into 9C2. Right. This value comes out to be a 36 into 36. Okay, 36 into 36 is 36, 9, 18, 36, 6, 9, 12. It is just 1296. Okay. So there are just 1296 rectangles on a chessboard. Right. Next part of the question, how many of them are squares or how many squares are there on the chessboard? Let me not write anything. I'll wait for you to respond. See, I would suggest you to follow the basics. Of course, there is a formula, but we'll go stick by the basics. So how many dimension one by one square will be formed? That is 8 into 8, 64. How many dimensions 2 by 2 squares will be formed? 7 into 7. How many dimensions 3 by 3 square will be formed? You'll say 6 into 6. And let's say I finally come to dimension 8 by 8 square. That will only be 1 square, which is the entire chessboard. So all you need to do is add 1 square, 2 square, up till 8 square. Isn't it? So what is the sum of squares of natural numbers from 1 to 8? What are the squares? What are the sum of squares of natural numbers from 1 to n? What is this formula? We did it in the chapter sequence series for questions, right? n into n plus 1, into 2n plus 1 by 6. So it'll be 8 into 8 plus 1, 2 into 8 plus 1 by 6. This is 8 into 9 into 17 by 6. This comes out to be 204. So there are 204 squares on a chessboard. Clear? Okay. So next part of the question is how many non-square rectangles are there? So this answer is 1296. This answer is 204. So this answer is 1296 minus 204, which is approximately 109, not approximately exactly 1092. So 1092 non-square rectangles are there on a chessboard. Okay. Now the fourth question, which was about, I was about to give you, but I stopped myself. Now I'm going to give you that question. But before that, do you have any questions, any concerns with whatever we have done so far? Okay. Now my question to you is, let's say there is a small insect, okay, which is sitting over here. This insect can only move along positive x-axis or along positive y-axis by one unit each. Okay. Do you remember we had a question in probability where there was a person who was trying to go to office and he can only move one unit? Remember that? Okay. Same as this insect also, this insect can move only one unit right or one unit up. It cannot go left. It cannot go down. It has only potential of moving right and up. Tell me if this insect has to reach from this position to this position, how many different paths can it take? So let's say it has to reach from position A to position B. How many paths? So that is the fourth question. How many paths are you? How many paths can the insect take to go from A to B and can only move, this guy can only move one unit in this direction or one unit in this direction? Yes. So the answer to this question is 16 C8 and that is something which I already discussed while solving the problem last week. So basically you have eight rights. Okay. And you have eight ups. You need to create a 16 alphabet word from eight hours and eight use. The number of ways in which you can create will basically decide the number of paths he can take or the insect can take to go from A to B. So it is 16 factorial by eight factorial, eight factorial, which is 16 C8. So this type of question also is asked quite a lot in the competitive exams. Is it fine? Any questions? So, insect, okay, see, insect, how will it go? Insect will either move right or upright. So let's say one way the insect can go from A to B. See, it has to ultimately move eight units, eight units up. See, sharply, you can be given a lot of ifs and buts in the question, but you have to apply your mind to solve it. It cannot be given, everything cannot be given to you as a theory. So one question, I think there was in KVPY where they said that the insect cannot go above the diagonal. What will happen in that case? That is called good path, actually. That's called a good path problem. Okay, you have to apply it. Another case that they had given was there is a hole which is created over here, something like this. They punched a hole like this. Now tell me how many ways can the insect go from A to B. So all those situations you have to think. Okay, but let me address Setu's question. Setu was asking, how did you get this? Now see, Setu, he has to take eight moves to the right direction and eight moves to the up direction to reach that point. Ultimately, he has to take eight R's and eight U's to reach from A to B. Now you can arrange it in different ways. So example, he can take R, R, U, U, U, U, R, R, U, U, R, like that. So basically, you can arrange it in different ways. This is one path. So this is one such word you can make from eight R's and eight U's. Isn't it? So every word that you make from these R and U's will decide the path. So how many such words can you make from eight R's and eight U's? Just recall your permutation combination skills. If you have 16 alphabets in which eight are identical of type one and eight are identical of type two, then how many words can you form? A 16 factorial by eight factorial, eight factorial. So the number of words you can form is the number of parts you can create. Is it fine? Is it fine? Okay. So now keep thinking about the complexities which I was discussing with Sharduli. If let's say the insect is not allowed to breach this, let's say I'm just going to connect here. Let's say the insect cannot go above this line. So he has to take such a path where he cannot cross this line. Then how many paths can he take? So that will basically become a case which is slightly more complicated. Okay. We call it as a Catalan's number. Okay. That's a higher level combinatorics. If you have done your combinatorics for PRMO, RMO, that is something called Catalan number formula. Okay. Search for it. You'll get to know the Catalan's number. Catalan's number gives you the number of good paths. This is called the good path where the insect is below the diagonal position. Okay. And also think if let's say if I create a gap somewhere, let's say if I remove this, that means insect cannot go through middle of this. Means there is a well kind of a situation created. So I've dug a well over here. Then how will the insect go? So think of those situations as homework. Okay. And do let me know if I can help you. So the last part of this topic is going to be derangements. I think I should be able to complete this before the break. And after the break, we can start with a new topic. What is derangement? Let us understand this. Derangement of N objects. When I say I am deranging N distinct objects, it means placing all the N objects in places not designated for it. Notice this. Write this down first. So when I am saying I'm deranging N distinct objects, it means I'm trying to place these N objects in places which are not designated for it. Let me give you a simple example. Let us say there are four letters. And these four letters are supposed to go in these four envelopes. Let's say this is envelope E1, envelope E2. I don't know whether you people have seen letters like this which are written, which are placed in envelopes. Nowadays email time has come. People have stopped writing letters. So let's say you belong to the old school of thought and you decide to write four letters to four of your friends. So letter one was supposed to go inside envelope one. Letter two was supposed to go to envelope two. Letter three was supposed to go inside envelope three. And letter four was supposed to go inside envelope four. How many ways can you place these letters such that no letter goes in its right envelope? That is the meaning of deranging these four objects. Clear? Situation is clear. How do we do it? That's a different thing altogether. We'll talk about it. But is the statement that I'm deranging four letters here clear to you? That means putting these letters in such positions or in such envelopes so that no letter goes into its designated envelope. Clear? Now various type of questions are framed. One type of question will be framed is like let's say there are six couples, husband wife couples. How many ways can you arrange a mix, you can say mix singles between these couples such that no husband plays against his wife? That is also a case of derangement. I think some of you, one of you asked me that question and I told I have not yet covered that topic. So this is the concept that will help you to solve that question as well. So a lot of different types of questions. Oh, it was Siddharth. Sorry, I forgot the name. Yeah. So these are the type of questions that will be related to this topic. Okay. So deranging N objects means putting N objects into such places such that no object goes in its designated position. Clear? Now, coming to this question, how to solve this question? I will come back to this question once again. But let us try to figure out the formula for derangement formula for N objects. Okay, let's derive this formula. And for deriving this formula, we have to again fall back on our principle of inclusion and exclusion. Okay, we have to rely on the formula of inclusion and exclusion. All of you please pay attention, even though the derivation is not important, but you should know how the formula actually comes up. Okay. Let me take a scenario like this. Instead of four letters, let me extend the scenario to N letters. Okay, so these are N letters. And these N letters are supposed to go in these N envelopes. That means letter L1 was supposed to be put inside even, letter L2 was supposed to be put inside E2 and letter LN was supposed to go inside. Okay. Now, let me ask you this question. If there is no restriction, if there's no restriction at all, how many ways can you put these letters in these envelopes? So let's say I call you universal set, no restriction guess. So how many ways can you place these letters in these envelopes so that there's no restriction? How many ways can you do it? Absolutely, right, Nikhil? N factorial, isn't it? It's just like, you know, there are N seats, there are N people. How many ways can you make those N people sit on those N seats? So making a person sit is like posting a letter in a envelope or putting a letter in an envelope. Okay, N factorial means like, yes or no, everybody agrees with this? You have to agree, we are in the last leg of this chapter. Okay. Now, let me give a situation here. Let's say even is a situation where L1 goes in its designated envelope. Okay. So let's say even is an event where L1 goes in its designated envelope. Okay. Others, I'm not worried about whether they go or they don't go, I don't care. But L1, this has to go in even. So can somebody tell me if L1 goes in envelope even? By the way, I should not use even here and even here. Let me use a different name altogether. What should I name it? Any suggested name? Even, should I call it even? Okay, so let's say even is an event where the letter L1 goes inside even only, then tell me in how many ways can I, can I basically do these posting of these letters or putting of these letters in this envelope? Write your response on the chat box. Okay, Nikhil. Why only Nikhil is answering? Others, what did you ask? I said that I have already put this letter inside even, then how many ways can I post the other letters without caring whether they go into their right envelope or not? Right, Siddharth? Anybody else? Right, Setu? Okay, N minus one factorial, everybody agrees? Okay. So can I say that any AI, let's say event if I take where LIth letter goes inside its respective envelope, can I say that even for this case, even for this case, your number of ways will be N minus one factorial. Okay. Now, see, try to recall the principle of inclusion exclusion which I had taken in proving the distribution of N distinct objects into our different groups where blank group is not allowed. Okay, same procedure I am going to take here, similar procedure I should say, not same. So let's say A1 intersection A2 is an event where both L1 and L2 goes in their respective envelope. Okay, then tell me how many ways can I do this event? L1, L2, go into their respective envelopes. Other letters, I don't care whether they are going or not going, I don't care. Tell me how many ways can I do it? What's your answer going to be in this case? N minus two factorial, everybody agrees to it? So even if I take any, let's say Ith and jth letters such that Ith letter goes inside EI and jth letter goes inside EJ, the same formula is going to be valid. Am I right? N minus two factorial only. I don't want to take all the several cases. Can I say similarly, similarly, the number of ways in which the Ith and the jth and the kth letter goes inside their respective envelopes that will become N minus three factorial and so on and so forth. Okay, now the main thing, what do I want? I actually want that letter L1 should not go into its right envelope or should not go into E1 and L2 should also not go into E2 and L3 should also not go into E3 and so on. This is what actually I'm looking for. Yes or no? Which as per De Morgan's law is going to be N of, oh, I'm so sorry, now I've changed the name. Thanks, Nikhil, thanks for bringing that to my notice. Thanks a lot. Okay, so can I say as per De Morgan's law, it is the complement of this expression, which as per this formula becomes this. Okay, now this further, I can write it as summation of this minus summation two at a time. I hope you all recall this formula, then plus summation three at a time and so on and so forth. Okay, same approach. Okay, so now NU is N factorial. So let's start putting the values against it. This is N factorial. Now summation NAI means you need to add N minus one factorial, N minus one factorial. How many times? N times, right? So you'll have NC1 into N minus one factorial. Everybody's convinced? Yes or no? Tell me what will be this, the second term. Who will tell me what will be this expression? What should I write? Two at a time. How many two at a time combinations can be formed? NC2 and each one of them is N minus two factorial. You only told me no. See, N minus two factorial, N minus two factorial. So if you want any two of them to go into the right envelopes, first of all, those two can be chosen in NC2 way. So there will be NC2, N minus two factorials that will appear in this summation. Okay, so that will give us this. Similarly, the next term will be NC3, N minus three factorial. Then I'll have NC4, N minus four factorial. And wherever it starts, you know, starts becoming up. You know, let's say if I have to write it, if I have to write it, I will go, okay, let me write it in the next step maybe. So let's simplify this expression a bit. This term is n factorial by one factorial, N minus one factorial. And you're multiplying it also with n minus one factorial. Similarly, this term, NC2 is n factorial by two factorial, n minus two factorial, and you are multiplying it with n minus two factorial also. Similarly, this term is n factorial by three factorial and minus three factorial, and you are multiplying it with n minus 3 factorial also and this continues okay till you reach till you reach ncn ncn i will write it like this n factorial by zero factorial n factorial into zero factorial okay and the sign here will actually depend on n value is it fine any questions any concerns so far now see this gets cancelled with this this gets cancelled with this this gets cancelled with this this gets cancelled with this similarly all these terms start cancelling and if i take n factorial common i will end up getting one and i open the bracket here i'll get one minus one by one factorial i think i have reached the end of the slide here so much was left on top okay i'll just repeat this again in the next page note this down note this down i'll i'll just copy paste this and i repeat it once again don't worry i don't want to mess up the simplification process then copied everybody i'll write once again give me a second uh okay yeah so this sign will be this sign will be n plus let's say one because for even it is becoming negative for or it is becoming positive right i think did i write that n plus one in the slide before it what was it oh sorry this is n plus one because for even it is becoming negative right so i have to have an odd number here clear makes sense okay now see when you expand this you end up getting something like this i'm just removing the terms which i cancelled out now take an n factorial common if you take an n factorial common right from the first term you get one minus one by one factorial plus one by two factorial minus one by three factorial and so on not this time this time this will become minus one to the power of n plus two you can say or you can say back to it because it doesn't matter whether you write n or n plus two the power is going to be not affected because of that so note down this formula is what we call in short form as dn d with a subscript of n it means it basically tells you how many ways can you derange n objects n distinct objects taken all at a time okay so now coming to that special case of letters so that letter question four letters are there okay and you want to derange those four letters so as per this formula your answer will be four factorial by one minus one by one factorial plus one by two factorial minus one by three factorial plus one by four factorial okay if you multiply throughout you'll get 24 minus 24 plus 12 minus four plus one okay this is nothing but 13 minus four nine ways you can derange so there are only nine ways in which you can put those letters such that no letter goes in its designated envelope okay this nine is also basically called d four any question with respect to the derivation of the formula any question with respect to solving that question which I started my discussion with if yes do let me know okay all right so a few things that I would like you to remember just to save your time how many ways can you derange one object so there's one letter and one envelope how many ways can you place this letter such that that letter doesn't go into its right envelope zero is right so you can't derange one object okay how many ways can you derange two objects one very good how many ways can you derange three objects correct so basically you have to use this formula whatever I gave you are you getting my point okay very good how many ways can you derange four objects we have already seen it nine ways how many ways can you derange five objects please calculate and let me know I want you to remember these figures actually that's why I'm giving you these figures you have to remember this it saves a lot of time how many ways can you derange five objects five letters five envelopes in how many ways can you place them so that no envelope goes into its designated envelope yes anybody are you getting 44 anybody who's doing it are you getting 44 hey it's not a rocket science yeah it's five factorial one minus one by one factorial one by two factorial minus one by three factorial alternating plus minus sign as you can see here now you can see here that this and this term will always get cancelled out so you have to start from your second term so it's my factorial you can say 120 one by two minus one by six one by 24 minus one by 120 okay so multiply it so it's 60 minus 20 plus five minus one so it's 40 plus four 44 okay d6 will come out to be 265 okay so just remember I mean be honest you don't need it okay let's take a small question based on the same and then we will take a break is it fine any questions here all right let's take a small question a person writes letters to six friends and addresses the corresponding envelopes in how many ways can the letters be placed in the envelopes such that so that part one of the question says at least two of them are in the wrong envelope okay and part two of the question says all the letters are in the wrong envelope okay shardulik part two everybody will get it right yes anybody with the first part okay very good nikhil say two ones one minute fine fine I'm giving you one minute okay now to solve the first part of the question there are two ways to do it one is I would say not a smart approach another is a smarter approach so I'll start with not smart approach okay so what is a not a smart approach see the question says you want at least two of them to be in the wrong envelope correct so first you need to choose which two so out of six you need to choose the two letters which you want to go into a wrong envelope correct which means the remaining four have gone into the right envelopes right let's say l1 l2 l3 l4 l5 l6 l1 and l2 are chosen to go in the wrong envelope that means l1 goes to e2 and l2 goes to even rest all of them are going into the right envelope something like this see I'm drawing a diagram on your this thing so these are your let me just name it like this making boxes is time consuming oh sorry okay and these are the respective envelopes so I chose those these two letters to go into the wrong envelope these four have gone into the right envelope going in the right envelope means there's only one way to send them into the right envelope now these two can go into a wrong envelope in d2 a of course d2 is something which we already figured out is one we'll come back to that similarly if you want three letters to go into a wrong envelope it will be this similarly if you want four letters to go into a wrong envelope it is going to be this five this and finally all the six going into the wrong envelope is going to be this is it clear why did I write one here because one is the number of phase in which these guys will go into the right envelopes there's only one way e3 should go l3 should go to e3 l4 should go to e4 l5 should go to e5 and l6 should go to e6 but these two can be switched in one way that is d2 actually we'll come to that so this is nothing but 6c2 into d2 6c2 is 15 d2 is 1 I believe 6c3 is 20 d3 is 2 correct 64 is again 15 d4 is 9 65 is 6 and d5 is 44 and cc6 is 1 and d6 is 265 if you add it how much does it give you 15 plus 40 plus 135 plus 256 265 how much does it come out to be I'm sorry this is going to be 264 sorry 264 yeah how much it come out to be just add it 55 plus this is 190 190 and this is going to be 529 so that's comes out to be 790 yeah 790 okay now can I say this problem can be solved alternately in a faster way first of all if there is no restriction at all how many ways can I put these letters in this envelope six factorial can I say subtract from it where all the letters go into their right envelope that will also give me the answer now see here everybody please understand if if you're subtracting the one case where all the letters go into the right envelope it means you are basically left with those cases where at least one letter goes into the wrong envelope but remember exactly one letter cannot go into a wrong envelope if the other five have gone into the right envelope the remaining one has to go in its own envelope only so there is nothing like exactly one goes into a wrong envelope so understand it so when you remove this case of one you are left with those cases where you are trying to say at least one letter at least one letter goes into a wrong envelope right but I say this is as good as saying at least two letters this is equivalent to saying at least two letters go into a wrong envelope why because there's nothing like exactly one letter going into a wrong so at least one becomes at least two actually so at least two yeah at least two letters go into wrong envelopes am I right or not so subtracting the case where all the letters go into the right envelope will leave you with those cases where at least two letters are displaced isn't it so the answer is in one shot coming out from it any questions any concerns so with this we happily end this topic and we'll take a break also right now