 Okay. So I'll be starting with a combination with combination with identical objects, identical objects continued. So let's continue with that concept. Okay. Under this first thing that we are going to talk about is finding the number of devices, number of devices, or factors, you can say, or factors of a number n, number of devices or number of factors of n. Okay. Good afternoon, everyone who has joined in. Good afternoon. Yeah. So let me ask a very simple and straightforward question to you all to begin with this concept. How many factors or how many devices do you think the number 24 has? Tell me how many devices the number 24 has. All of you can do it on your notebook and tell me the result. Five. Five. Six. Six. Okay. Okay. Anybody else? Seven. Seven. Seven factors. Okay. Seven. Seven. Okay, final. Should I lock it? Yes, seven. Okay. So let's try out. What is the factor? What is the factor of everything? We know that. Two, three, four, six, eight, twelve, twenty-four itself. Remember? Twenty-four and one are called improper factors. So these two are called these two are called improper factors. Improper factors. Whereas the rest of the factors are proper factors. So these are proper factors. Okay. Okay. So your answer is not three, not five, not four. It's actually eight factors are there for 24. Right? Now this number was small. That's why you could actually sit and count it. Correct? But if let's say I give you a number like this, let me slide this up. How many factors does this number? 38,808 have. Right? How many factors or how many devices that this number have? How will I solve this problem? Anybody has any idea how to solve this or anybody knows how to approach with this problem? A rough road map? If not giving me the solution, at least tell me how to proceed with this? Any idea? Now let me tell you this concept is very similar to the concept of combination from identical objects. Now how you need to tell me? I'm inviting all the opinions. So please feel free to unmute yourself and speak your mind. In order to answer this, first try to address this cause. How would you actually find out the number of factors of 24 without actually listing them down like this? Of course, listing here becomes easy, but you cannot list all the factors of the number becomes as large as this one. So what is the approach that you are going to take? If at all you want to find out the number of factors of 24 without actually writing them down. Any idea? Just think on it for one minute and let me know. Sir, I guess we can prime factorize it and then find all the permutations of those prime factors. Okay, who is this by the way? Siddharth. So Siddharth has suggested an idea that let's prime factorize 24. Okay Siddharth, I'll do that as 2 to the power 3 into 3 to the power 1. What next Siddharth? I'm not sure. Okay, now you're actually done half the work. Find out the combinations of each of the prime factors. Find out the combination of each of the prime factors. Anybody else who wants to put forth his suggestion or approach? You're most welcome to do that. Now listen to my argument over here. My argument is if I want to make a factor, then that factor could be made of none or more 2s or none or more 3s. Yes or no? Let's say I want to make a factor like 4. You can see the 4 is the factor. This factor is made by choosing 2 2s and choosing no 3s. Getting my point. If I want to make a factor like 12, I have to choose 2 2s and 1 3. So any factor that you want to make from here, let's say I want to make 1 only. One can be obtained by choosing no 2s and choosing no 3s. Are you getting it? So my logic here is or my roadmap here would be to find out the number of ways of picking none or more. Watch out my word. None or more 2s and none or more 3s. Are you getting my point? Correct? Now if I pick up nothing at all by default that factor becomes 1. If I pick everything by default that factor becomes 24. Are you getting my point? So think as if there is a fruit basket in which there is a fruit called 2 which is present 3 times because there is a power of 3 on 2. And there is another fruit called 3 which is only present once. And you are asked to pick up none or more fruits from this basket. Now can you solve this? You will say very simple if I have to pick up none or more 2s I will pick up in 3 plus 1 4 ways and if I have to pick none or more 3s I will pick it up as 1 plus 1 that is 2 ways that is 4 into 2. Doesn't this give you the total number of factors that we have for 24? Right? 8 is the total number of factors. Now does this give you an idea how to scale this up even if you have a bigger number like 3 8 8 0 8. If yes you have understood please go ahead and solve this. If not please raise your hand. You can always raise your hand. You know how to raise your hand also right? Or you can just unmute yourself and ask me. Is that clear everyone? At least type clear if it is clear so that I know everybody is on the same page. I will repeat this once again for the people who have joined us late. If you want to count the number of factors of any number In this example I have taken 24. So just prime factorize 24. For example 24 is prime factorize as 2 to the power 3 into 3 to the power 1. Now think as if you have a fruit basket where there is a fruit called 2 and there are 3 such fruits in that basket. That means the fruit called 2 is present in number of 3. And there is another fruit called 3 which is only present 1 in quantity. And you have to pick none or more fruits from this basket. So the answer is 3 plus 1 into 1 plus 1 that is 4 into 2 answer is 8. In a similar way I would like you to attempt how many factors are there for this number 38,808. I am giving you 2 minutes your time starts now. Let me see who all have come. Ronak is there. Yeah Ronak you raised your hand all well. Sir is the answer 72 along with improper factors. Yes sir 7. In today's class we will follow a norm. Whatever answer you get you will privately message me so that others are not influenced by your answer. So let's follow this norm while we are having online classes. I don't answer to everyone just answer privately to me. I will just say right or wrong to you so that if you are right you can relax for some time. Or if you are wrong you can start reworking on it. Okay sir. By the way Sankin your answer is absolutely correct 72 is the answer very good. So I am sure everybody would have prime factorized 38808. 38808 can be fine factorized as 2 to the power 3 into 3 to the power 2 into 7 to the power 2 into 11 to the power 1. Hope nobody has made a mistake in this correct. So the total number of factors the total number of factors of this number would be nothing but 3 plus 1. Just add a 1 to the powers of these 5 factors 2 plus 1 again 2 plus 1 again 1 plus 1 that makes it 4 into 3 into 3 into 2. That's nothing but 18 into 472. Is that fine. If the question says how many proper factors are there you will just subtract 2 from it. So number of proper factors would be 72 minus 2 which is nothing but 72 which is nothing but 72. Okay fine. Now I have a slightly tweaked version of the question for you. Okay. Tell me how many even factors are there or even divisors are there for this number. Even divisors or factors whatever you want to call it. How many even divisors are there for 38808. Again reply to me privately. Just a slight tweaking of the question but I'm sure you'll be able to handle this. Okay. No Mayur that is not right. So when you're asking for even do we have to include the improper faction also the factors. See if the improper factor is an even factor for example here the number itself is even then you have to include it yes. Okay sir. Anurag no. By the way who has no Anurag that's wrong. Trippan you are correct. Shankin you are off by 1 I guess. I don't know why you chose to subtract a 1 but you are off by 1. Skander no that's not correct Skander. One person has so far given me the right answer. Yes correct. It should be that yes. Who would like to take a try how many even divisors or even factors are there for this number. That's correct Ananya. Good so three people have answered correctly so far. Okay. I'm waiting for. No Anjali. No Siddhartha. One more answer I need. Narasimha that is not correct. Okay everybody please have a look at my explanation. See if you want to have an even divisor you must at least pick one two in your factor. Correct. That means the fruit two that you have in your basket you don't have an option to pick none from it. Right. You have to pick at least one or more than one. Okay. So you have to pick at least one fruit which is the fruit called two. Correct. So that that three plus one that used to do now you will not do that plus one because that plus one was to include the none case. Right. Here you don't have the option to choose no twos. Right. So you have to pick at least one two so that is three ways because you can pick one or you can pick two or you can pick three. So three ways all together. But for the others my dear you will have two plus one because you can pick none or more threes. Right. So two plus one two plus one one plus one getting my point. So the answer will be three into three into three into two that is nothing but 54. So there are 54 even devices or factors for this number. Needless to ask you how many odd devices would be there. You would simply subtract the total number of devices with the even number. So 72 minus 54 18 odd devices are there 54 even devices are there. Right. So those who answered incorrectly now you know where you went wrong. Okay. So these kind of tweaking can be possible in such questions. Right. Now what I will do is I'll quickly generalize this. I'll quickly generalize this. I will quickly generalize this. Okay. If there is a number N. Okay. Which is prime factorized as P1 to the power alpha one P2 to the power alpha two P3 to the power alpha three. Da da da da da. Till PK to the power alpha K. Okay. Where P1 P2 etc are prime factors. Okay. And alpha one alpha two etc are the exponents of those prime numbers. And the number of devices of that particular number N will be nothing but alpha one plus one alpha two plus one dot dot dot dot. Till alpha K plus one. This is the total number of devices. Okay. And if you want your proper devices, number of proper devices, number of proper devices, you just have to subtract the two from your answer. Is that fine. So this concept is a very hot concept. It is asked a lot in various competitive exams, not only in JEE but other regional entrance exams as well. Okay. Now this concept doesn't end over here. I'm going to go a slightly more deeper into this and ask you now this question. Find the some of the devices. Find the some of the devices. Right. Of 24. Okay. So now I'm going to talk about the concept of some of devices. Some of devices. Let me name the topic as some of devices. So I'm starting small now. I'm first focusing on 24. Find out the sum of the devices of 24. Now I'm sure you would sit and add them up. You'll say, okay, one plus two plus three plus four plus six plus eight plus 12 plus 24. Okay. Let me give you a count for this. 24, 24, 24, 48, 48 plus 12, 60 is your answer. Right. But how do you get this without actually sitting and writing those factors and adding them up. Okay. So let's say the number becomes big like this. Right. Let's say my question is find the sum of the devices of devices of three, eight, eight, zero, eight. The same old number that we had in the previous question. How will I do this? Try to figure out a technique or a method by which you can solve the problem of finding the sum of devices of 24 without actually counting them. If you have a method, you can share it with me. You can unmute yourself and talk. The answer to this is hidden in that previous question or previous concepts itself that is counting the number of devices. Anybody has any idea? Anyone? Skanda, Ananya, Tridib, Siddhartha, Shaumik, Shristi, No, no, no, no, no, that's not the correct answer. Let us first figure out the methodology by which we can sum up the factors of 24 first. Okay. Now, how is 24 factors generated? Let's have these eight cases. Basically, you're trying to make a factor by choosing none or more of twos and threes. Let's say I choose nothing of these both. That means I get a factor one. This is covered. If I choose one of two and no threes, I end up getting a two. Yes or no? If I choose two twos and no threes, I end up getting a four, which is this. Okay. If I end up choosing all twos and no threes, I end up getting eight. Yes or no? Similarly, if I choose no twos and one three, I get a factor three. Correct. Yes or no? If I choose one two and one three, I get a factor six. Correct. If I choose two twos and one three, I end up getting a factor of 12. Correct. And if I pick up three twos and one three, I end up getting a factor 24. Correct. So these factors that you see are generated by the following, you know, combinations of these powers of two and three. Okay. When you are adding them up, when you are adding them up, you would realize and you would appreciate the fact that you actually ended up adding this with this. Yes or no? See, when these two got multiplied, you got this factor. When these two got multiplied, two to the power zero, three to the power one, you got the factor of three. Correct. When this thing got multiplied, two to the power one, three to the power zero, you got this factor. Okay. When this got multiplied, you got six. Correct. So basically, when you take up these terms and multiply them together, you'll start realizing that all these factors have started featuring up. Correct. That is not right. Okay. In other words, what I'm trying to say my dear students is that if you multiply these two terms, let me just erase these parts. If you multiply these two terms, it will give you all the eight factors which are getting added up. In other words, this itself gives you the sum of all the factors or all the devices of this number. Right. Let me just complete this and show you how you get the sum of 60 from this. So this is nothing but one plus two plus four plus eight. By the way, you can also sum this up as two to the power four minus one. Okay. So you know a GP. Geometric progression. Right. I'm sure you would have done in this autonomous course in class 10. Okay. Those who don't know a GP looks like this. A, A, R, A, R squared, A, R cube, all the way till A, R to the power N minus one. Correct. Where A is the first term and R is the common ratio. Correct. The sum of a GP to N terms is given as A, R to the power N minus one by R minus one. There I'll write A is the first term. A is the first term. R is the common ratio. A is the first term. R is the common ratio. Okay. Have you all done geometric progression in class 10? Yes or no, guys. I would like to know whether, yeah, I'm sure. No, sir. Why, sir? No, sir. We didn't learn. Oh, okay. It doesn't depend upon your understanding of geometric progression to solve this. Okay. Tushar sir has taught you. Okay. I'm sure Tushar sir would have taught you last year. Right. So now, even if you have not learned it, no problem, you can actually literally add it up. So one plus two is three, three plus four, seven, seven plus eight, 15. 15 is nothing but two to the power four minus one. Okay. Anyways, this itself is one plus three, which is four. So the answer is 15 into four. And here goes the answer. So we got the answer 60, which we had already got by literally adding them up. Right. Now, if you have to scale this concept to find the sum of the devices of 38,808. Now are we ready to do that? If yes, can somebody quickly tell me the answer within one minute? What would be the sum of the devices of this number? Time starts now. Please answer to me privately. Don't put it to everyone. Some of all the devices, that is my question. By the way, those who want to know how it was fine factorized, it was fine factorized as two to the power three into three to the power two into seven to the power two into 11 to the power one. Now take a hint from this methodology and try to scale this up to solve this question. This is called application. Okay. If you're looking small and you scale this up to solve a bigger problem, that is called application. Jay is looking for people who can apply. In fact, all entrance exams look for people who are willing to apply the concept. That's absolutely correct. Brilliant. Oh, no, no, Krishna, you are shot by a big margin. The number itself is bigger than this. How can the sum be lesser than the number? The number itself is 38,000. Come on. Okay. Can you take the effort to simplify that, Siddhartha? Anand, you are absolutely correct. Shankin, you are off by 12. I don't know again how. Srishti, you are correct. No, Vibhav, I'm afraid that's not the answer. Vibhav, can you try again once, please? Guys, please participate. Those who are just sitting like this. I would invite you to put forward your answer. Any ways you are answering me privately. Let it be wrong. No, Skanda, that's not the answer. Yes, Shankin, that is correct now. Okay. Time up. Let's discuss this. Correct, Vibhav, correct. No Aditya, that is not correct. Oh, can I see it? See, what I did was, here the logic is, let's say the factor 2 here was, this was prime factor as 2 to the power 3, 3 to the power 1. So basically what you do is, every prime factor you start from 0 and take it to the highest number that it is occurring in that particular number. For example, 2 was occurring 3 times, so I'll start from 0 and go all the way till 3. 3 is occurring once, so I'll start from 0 and go all the way till 1. Okay. So in a similar way, I'm going to attack this problem as well. So what I'm going to do is, I'm going to take this number 2 first, start it from 0, go all the way till 3. Okay. 3 to the power 2, also I'll start all the way from 0, go all the way till 2. Okay. 7 also, I'll start all the way from 0, go all the way till 2. And 11, sorry, I have to shift my screen a bit, 11, 11 to the power 0, 11 to the power 1. Right. I'm sure you can add this pretty quickly. This we have already done, that's 15. Okay. This guy is going to be 1 plus 3, 4, 4 plus 9, 13. This guy is going to be 50, 57. And this guy is going to be 12. Okay. I'm sure you can multiply this. Let me use my calculator in order to save time for everybody. But remember, you'll not be provided with any calculator in the exam, so you have to do your calculations on your own. And my calculator throws out this number, 3, 1, 3, 3, 8, 0. So those people who have got this answer, there are some more people who are replying me afterwards. Ronak, no, Ronak, Ravi, Siddhartha, Krishna, Anjali. Yeah. Please, please see my working now. I'm sure you would have understood by now. No problem, Ronak. I always appreciate people who try. Did it be wrong? No problem. Okay. Yes. So this is going to be the way to solve this problem. So if I ask you, what is the sum of the proper devices? What do you do? If I ask you the sum of the proper devices, all you need to do is subtract the number and 1 from it. Okay. That's going to be 1, 3, 3, 8, 0, minus 3, 8, 8, 0, 8, minus 1. That is going to be 9, 4, 5, 7, 1. That is the sum of all proper devices. Does that make sense to everyone? Clear. Please type clear on your screen if this approach is clear to you. This is one of the most important concepts. Let's not move forward till we have understood this concept. Okay. I have next question coming up for you. And here it goes on your screen. In how many ways can the number 1, 0, 8, 0, 0 can be resolved as the product of two factors? Okay. Before you start solving this one, please let me conclude the previous theory. Okay. So guys, just excuse me and I'll just conclude the previous concept. So I'll generalize this. I have not generalized this yet. So kindly allow me to generalize this. Then we'll go back to that question. Let's generalize this concept. So if you have a number which is prime factorized as p1 to the power alpha 1, p2 to the power alpha 2, p3 to the power alpha 3, till pk to the power alpha k, then we say that the sum of all the devices is given by, sum of all the devices is given by p1 to the power 0, p1 to the power 1 all the way till you reach p1 to the power alpha 1 times p2 to the power 0, p2 to the power 1 all the way till p2 to the power alpha 2 and this trend continues till pk to the power 0, pk to the power 1 all the way till pk to the power alpha k. Okay. Those who would be interested in knowing its simplification, its simplification is p1 to the power alpha 1 plus 1 minus 1 by p1 minus 1. You can note down this result as well. It will save a lot of time for you later on. p2 to the power alpha 2 plus 1 minus 1 by p2 minus 1. Okay. And this multiplication continues till we reach the last term over here, which is pk to the power of, pk to the power of alpha k plus 1 minus 1 by pk minus 1. Okay. This result, don't be surprised. This comes from your GP, comes from the sum of a geometric progression, which anyways you're going to learn in the chapter series, sequence and progressions later on with me. Okay. So now we are ready to go to the next concept, next concept. In fact, next question that we had given over here. In how many ways can you write 1 0 8 0 0 as a product of two factors? Now what is the meaning of that? Two factors means let's say one way would be to write it like 108 into 100. So these are the two, two factors. Let me tell you 100 into 108 is also counted as the same way of writing it. Okay. So don't doubly count this. They are the same way of representing the same thing. Okay. Let's see somebody has replied to it. Yes. Sure. What I did. Okay. Let me just go back to the previous board once again. See, we had already discussed the previous example that you are going to start with the prime factor from power 0 all the way till you reach alpha 1. Yes or no? Okay. So that's what I'm doing by generalizing it. So remember I had a problem 24, right? 24 was 2 to the power 3 into 3 to the power 1, right? To find the sum, what did we do? 2 to the power 0, 2 to the power 1, 2 to the power 2, 2 to the power 3 whole multiplied with 3 to the power 0, 3 to the power 1, right? Correct. So in the same way I'm now generalizing it. I'm making it scalable. So if a factor is going, yeah, if a factor is having alpha 1 power will start from 0 all the way till alpha 1 and keep doing this and multiply them all. Does it make sense to you? Arpita? Yes, sir. Okay. Good. Good to know that. Let's move on to the problem. Yes. Anybody who is ready with the answer? How many ways can you write this number as product of two factors? Ananya, no, that's wrong. Again, start small. Start small. You'll come to know the pattern. No Ranganath. That is also not correct. That is also not correct. No Arpita. How? No Shamik. No Tripan. Okay. No Arpita. Okay. We'll come back to this problem after having seen the example of 24. Okay. So let's say I want to know in how many ways can 24 be written as a product of two factors? Okay. Now let's try to count the number of ways. 1 into 24 is one way. Correct. 2 into 12 is another way. Correct. 3 into 8 is a third way. Correct. 4 into 6. Right. Next, you would feel like writing 6 into 4, but remember 6 into 4 is already covered under 4 into 6. Correct. Next, you would feel like writing 8 into 3. 8 into 3 is already covered under 3 into 8. Next, you'll feel like writing 12 into 2. That is already covered under 2 into 12. Correct. Next, you'll feel like writing 24 into 1. That is already covered under 1 into 24. So the answer here is you have just 4 ways. Okay. Does this ring a bell in your mind? Sisti, you are correct. Awesome. You were quick to bounce upon it. This is a simple example. Very good, Sisti. Others, does it ring a bell to you that what is it calling you? Sir, the number of ways to represent it as product of 2 factors is half the total number of factors. Okay. Now who is this, by the way? Anurag, sir. Anurag, okay. Anurag, you have concluded it very well, but remember there are some exceptions to this. For example, let's say 36. First of all, let us write down all the devices of 36. Okay. What are the devices of 36? Can we all correct Krishna? You are correct. Okay. All of you please pay attention here. 36 has factors 1, 2, 3, 4, 6, 9, 12, 18, 36. Can you count how many of these are there? 1, 2, 3, 4, 5, 6, 7, 8, 9. Now if you say Anurag, your answer for writing 36 as a product of 2 factors is 9 by 2, doesn't it become absurd because 9 by 2 is a fraction. 9 by 2 is a fraction. Right. So what went wrong? Okay. That worked for this one, but did not work for this. Anurag, can you tell me why it didn't work for this? Anybody? Open to anybody. Why? I think the number of factors did not work for 36. 36 is a perfect square. Absolutely. So when a perfect square comes in, that approach of halving the total number of factors of a number to find the number of ways in which you can express as a product of 2 factors fails. Right. Why? Because I will show you right now. 1 into 36. Correct. But it has another factor, 36 into 1. That's why there's a pairing happening here. Okay. So I'll add it down over here. This pairs up with 36 into 1. Okay. There's a pairing happening between these two. Then there's 2 into 18. Then there's something called 18 into 2. So there's a pairing happening between these two fellows. Then there's 3 into 12. 12 into 3. So there's a pairing happening between these two. Correct. There is 4 into 9, and there's 9 into 4. So there's a pairing happening between these 2. Now when there is 6 into 6, there is no pair to it. Okay, so when you see this, basically what you're doing, you are writing every factor at least once over here, right? You see it? Okay, but here you're repeating the same factor 6 with itself. Okay, so the number of phase in which you are going to write 36 as a product of two factors is just going to be 1, 2, 3, 4, 5. Okay? Now, how can we generalize such a thing when you have a perfect square coming up and when you have not a perfect square coming up? For not a perfect square, I'm sure you would have found the trick. When n is not a perfect square, not a perfect square, so what you're going to do is you're just going to do the total number of factors, the total number of factors divided by 2. This is the number of ways of writing it as a product of two factors. Product of two factors. Right? But if n happens to be a perfect square, it's absolutely correct. Shaumik, Kirtan, Siddharth, you're all giving the right one. Yeah. So if it is a perfect square, okay? The number of phase would be total number of factors. Okay? You can say minus one by two plus one, which is actually a stupid way of saying total factors plus one by two. Both are the same thing actually. Okay? Why I subtracted a one is because that repeated factor I subtracted other factors are always repeated. Okay? So that's six and six two factors which are okay. That six factor is a repeated factor to make a 36 that I subtracted. There's others will be paired up. So I divided by two and ultimately added up a one to account for that six. This itself is simplified to total number of factors plus one by two. So just remember this that would be sufficient. So the number of phase of this is a total number of phase of writing a number as a product of two factors. If the number happens to be a perfect square. Is that clear? So now answering that previous question of mine. So one, okay, let me go back to the previous question. So page number four, it was yes. Yeah. So coming back to this question. One zero eight zero zero. I'm sure you would have prime factorize it by now. It is two to the power four. Three to the power three into five to the power two. So total number of factors or total number of devices for this will be four plus one. Okay. Three plus one, two plus one. Now my question here to all of you is how do I know whether it's a perfect square or not? Perfect square will give an odd number of factor. That's one thing. Another thing is that if a number is a perfect square, you would realize that all the powers of these prime numbers would be even. Don't you think that correct? A perfect square will have all the exponents of those prime numbers as even okay. So here three doesn't have an even power. It has a power of odd number. So it is not a perfect square. It is not a perfect square. If it is not a perfect square, as Krishna also lightly pointed out, the total number of factors will come out to be odd for a number which is a perfect square. So here my number of ways in which you can write it as a product of two factors would be nothing but this divided by two. So answer is five into four into three divided by two. Two to one answer is 30. That is the right answer. Is that fine everybody? Are you able to connect with this concept? Okay. So let me ask you another one. Please stop me if you have not understood anywhere and please write clear. Keep writing this clear. Even though if I forget to say it, please write it down. Okay. Here comes the next question in front of you. Find the number of devices of this number. Thankfully this number has already been given to me as a prime factorized form, which are perfect square. Find the number of devices of this number, which are perfect square. By the way, it's not completely fine factorized. They have ticked you. This is nine to the power 11. So be careful. Okay. Let's see who answers this first. Time starts now. I'll give you three minutes to do this. Please reply privately to me. Shamik. No, that's not correct. No, Krishna. That is also not correct. No analog. That's way too much. You have to tell those factors which are perfect square themselves. Getting my point. For example, four could be a perfect square factor for this. Right. No, Shristi. No analog. That's not the right answer. No trippan. That's also wrong. See, this is a streaking of the same concept. Okay. I'll just tweak the same concept. Siddhartha, no, Shankin, no. Okay, guys. Good to see you all working it out. No Anjali. That's not correct. Okay. Now all of you please listen to me. See, if you want a factor to be a perfect square. Okay. Remember that factor should be made up of even number of prime numbers. For example, four. Okay. Four is made up of two to the power two. And all the other factors would be zero, zero, zero. Right. By the way, this number has not been written in a proper way. Okay. Just to confuse you, they have written nine over here. So what I'll do is first I will make myself. No pratham. That's not correct. First of all, I will write this as two to the power three, five, five to the power seven, seven to the power nine. And again, this is three to the power 22. Yes or no. Right. So altogether I have two to the power three, three to the power 27, five to the power seven, seven to the power nine as our number. Correct. Okay. Now see if you want to make a divisor. Okay. Let's say there's a divisor called K. Okay. This device there, whatever it is made up of, let's say two, three, five, seven, it should always have even number of powers on top. It should always have even, even, even, even. Do you all agree with me or not? Yes or no. Yes or no. Just type yes. Shunk in my God. That's correct. Wow. Awesome. Shunk in has given the right answer. Well done. Shunk in. Okay. So of all of you agree with me that there has to be an even choice of power, right? Very good. Yes sir. Now two to the power three, how many even powers I can pull out? I can either pull out two to the power zero or two to the part two only. Correct. Yes or no. From this guy. Yes or no. That means only two ways. Yes or no. They're two to the power one. Sorry. Sorry. That's, that's two to the power zero and two to the power one. No, no, no. Even even powers, even power is, what is the even power that you can go to? Yes sir. Zero and two only, right? Yes sir. No problem. Now three to the power 27, you can pull out three to the power zero, three to the power two, three to the power four, three to the power six, three to the power eight. Oh my God. 10, 12, 14, 16, 18, 20, 22, 24. 26. Oh my God. How many are there? One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14. So 14 ways you can select the powers of three. Okay. Now five. Let's check zero, two, four, six, four ways to do it. Okay. Seven will be zero, two, four. Some people have started answering it also. No, no, no, no, no, Anurag, still you're wrong, no Krishna that's also wrong. Six and eight. That is how many five I guess. Yes or no. So the total number of ways you can end up making an even device. Sorry, perfect square divisor is nothing but the product of these numbers two into 14, into four into five. that's nothing but 28 into 20 answer is 560 ways guys remember how is to make the factors how is to count the total factors either you could put a factor 0 number of times or 1 number of times or 2 number of times still that maximum amount you can take right so that is to give you alpha 1 plus 1 but here that same concept is slightly tweaked that's why Jee doesn't want its candidate or aspirants to come up memorizing a formula you have to know the logic from which that particular formula is evolving so here I have only the liberty to pick up 2 to the power 0 or 2 to the power 2 that means only 2 options are there when I talk about picking up various powers of 2 similarly there are just 14 options when it comes to picking up so even powers of 3 there are only 4 options for picking up even powers of 5 and again 5 options for picking up even powers of 7 so altogether 560 is the total number of ways in which you can pick up even devices from this number is that clear please type clear great guys so now we are going to talk about a different concept which is a very important one so all of you please listen to me very very carefully the concept of division of distinct objects into groups division of distinct object into groups please note that when you are saying groups I mean non blank groups non blank means there should not be any group which is empty correct there should not be any group which should not contain any object is that fine so let us talk about this concept I will start this concept by asking you a question okay let's see how many of you are able to answer this okay question is there are 6 students or 6 people let's say you want to make 2 groups out of it one having 4 people and other having 2 people how many ways can you do that I hope the question is clear you have 6 people you are making 2 groups out of it 1 containing 4 people other containing 2 people right absolutely correct so 3rd one says 6 C 4 that's very obvious right because the moment you select 4 people for the first group the other 2 people are already selected isn't it yes or no right but let me tell you this answer is valid when the order of this group is not important what do you mean by order of the group is not important means I am not giving any name to this group I never said that this is my group A or this is my group B like this I just said you have to split it up as 2 separate groups right I never gave any name to the group I never gave any order to the group then this answer is perfectly correct and this if you simplify it it becomes 6 factorial by 4 factorial 2 factorial right okay now you would think sir this is a simple thing right but let me tell you that it can become complicated at times okay let me take the same question now now you have let's say 6 people you want to make 3 groups out of it one having 1 other having 2 and another having 3 again order of the group is not important then how many ways can you do that please message me privately Aditya how 6 3 4 plus 6 p2 this is defying all logic my dear that's not correct 6 4 this 2 is 4 people other 2 are already chosen by default then already 2 groups are formed 6 1 5 c 2 3 pun says that's absolutely correct correct correct so 2 people have answered so far and both are correct Xiaomi is also correct why plus Anjali may I know that why are you adding it adding means your event is completed by doing just any one of the activity that means even if you fix one person your job is done is that the case I don't think so absolutely correct Krishna, Vibhav, Ananya you are absolutely correct so various people have answered this correctly the answer is 6 c1 into 5 c2 into 3 c3 okay many of you have answered it in your own way but remember when you open this particular you know binomial coefficient what you are going to see is 6 c1 is 6 factorial by 1 factorial 5 factorial 5 c2 is 5 factorial by 3 factorial 2 factorial and 3 c1 by the way it's just one okay so you don't have to write it so even if you don't write this that's going to be fine because anyways 1 at the end of the day because whatever is left off that default becomes another group let me simplify this if you simplify this you are going to see something like this 6 factorial by 1 factorial 3 factorial sorry 1 factorial 2 factorial 3 factorial okay now this is a pattern to me actually okay the pattern is or in fact the theory is let me generalize this if you have m plus n plus p plus r plus dot dot dot objects okay remember they are distinct objects I have already stated this as a heading to this particular topic that I am talking about distinct objects if you are going to put these objects into m n p r dot dot dot groups like this remember here all the groups are of unequal size now many people would ask would it matter if the groups were of equal size or some of the groups were of equal size yes my dear it's going to matter I am going to take that up a little later okay so please note this down all of you if you are going to if you are going to split m plus p plus r plus m plus n plus p plus r dot dot dot dot how many objects are there remember it should be distinct into various groups let's say one group has m number of objects another has n another has p r like that remember groups are of unequal size and order is not important and order is not important order is not important order is not important means you are not giving any number to this group you are not calling it as group 1 group 2 group 3 or you are not calling it as team A team B team C like that then the number of ways to do that will be equal to m plus n plus p plus r whatever it goes on till factorial divided by m factorial n factorial p factorial r factorial dot dot dot dot okay so as you can see the previous two questions have fallen under this category and this is the result okay now I would like to go back to this question the very first question that I asked you tell me if the order becomes important if the order becomes important then how many ways can you split the six people into two groups privately answer me 64 was already there 64 was already there what will be done additional to 64 if the order becomes important my dear absolutely correct Shomik very good guys this should this is easy come on everybody should answer this okay correct all of you are answering it correctly well done okay guys the answer is going to be 64 into 2 correct in this case order is important let me write it over here if order is important you are just going to shuffle these people in these two groups right so 64 and these two groups can be inter arranged in two factorial or two ways no no no Anjali multiply it with the factorial of the number of groups okay that's what I have done over here getting my point here also if the order becomes important what will happen if the order is important remember there were three groups here so what will you do six factorial by one factorial two factorial three factorial into three factorial see why am I multiplying with the factorial see let's say I talk about the previous example in previous example let's say six people sorry four people were in group A and two people were in group two correct can we not switch the number of people in these groups now two people in group A and four people in group B so two ways here so two into two factorial is justified by that here let's say A, B, C are the name of these groups so one here two here three here one here three here two here then one sorry two here one here three here two here three here one here then three here one here two here three here two here one here right so six more ways will be added up sorry six more ways will be multiplied up so this is why a factor of three factorial is coming up are you getting my point out of you so if the order becomes important let me generalize this now if the order is important okay let's say I am making K groups okay right now I have not written how many groups I have made over here but let's say if I was making K groups then the number of ways would be m plus n plus p plus r dot dot dot depends upon how many objects there are by m factorial n factorial p factorial r factorial dot dot dot dot into finally K factorial you have to do my dear friends let me tell you are you getting this point is it justified why I am multiplied with K factorial okay can I ask you yes sure Aditya right sorry trippan Shashank Shashank why I am multiplied with K factorial is because those number of objects in various groups can change hands right see let's say let's say I give you a question how many ways can you split five as two groups of two and one group of three okay now if I say order is important let's say this group is called A and this group is called B correct so you can do two three and three two both possibilities are there so additionally your answer of five C2 into three C3 got multiplied with two right this is because they can be switching off the number of objects for the same combination between these two groups correct let's say these five people are trippan, shawmek Aditya, Ronak and who else let's say Aditya C okay so let's say team A is made up of let's say trippan and shawmek team B is made up of Aditya, Ronak and let's say Aditya C now they can interchange their position also that means now trippan and shawmek has gone to team B Aditya, Ronak and Aditya C has gone to team A correct so two more possibilities are there for the same breakup correct here order was important okay so it is not merely splitting up that we are looking at we are also looking at which team has what split up getting my point great to know that so let me ask so are you guys tired of studying chemistry when there is a lot of break it makes you study more and more and more isn't it okay I have a question for you all let's see whether you are able to apply whatever you have learned in how many ways can 12 different balls be divided between two boys one receiving 5 other receiving 7 let's answer the first part of the question answers are welcome for the first part your time starts now I'll give you 1 minute I'll give you 2 minutes for the first part that's not right that's correct please type in in terms of C rather than writing in terms of a number Siddhartha can you type it in terms of NCR something like that or at least factorial Arpita no dear that's not correct Arpita, trippan you are correct Shamik you are also correct absolutely correct Krishna absolutely correct Siddhartha no Shrishti that's not correct now let me handle this chaos please note that you are distributing these balls to two boys when you say you are distributing it between some people it automatically becomes the case where order is important order is important which one receives 5 and which one receives 7 becomes important you are not just making 2 groups out of them here you are giving it to 2 boys when you are assigning it to people order becomes important Kirtana why a plus 7 C7 is it not into 7 C7 oh thank god if you make such mistakes that gives me a shock that even the basic concept of counting is not clear so don't give me that so the first part of the question your answer will be factorial by 5 factorial 7 factorial into 2 factorial because when you are giving it to boys implies order is important so please note in the question they will not say hey order is important here order is not important it is up to you to interpret that getting my point okay next part of the question also in how many ways how many ways can these 12 balls be divided into groups of 5 4 and 3 balls respectively correct Shamik correct correct Krishna Vibhav no that's not correct correct Shruti one second yeah that's correct Siddhartha that's correct Anjali no that's not correct Arpita no not correct Ajay no not correct see now when you are just putting into groups that means order is not important that means order is not important just groups means order is not important you are just segregating it as 5 4 3 balls that's it you are not assigning it to anybody so nobody is going to tell you again I am repeating this I am not going to tell you whether order is important or not so you have to figure out on yourself okay so this is going to be the answer that's it that's it correct Vibhav correct Rannak okay no multiplying it with 3 factorial and all okay had the question been how many ways can you distribute 12 boys sorry 12 balls as groups of 5 4 3 and assign it to 3 boys then into 3 factorial will feature in else it will not else it will not getting my point okay next question let's try it out clear guys let's try it out let's try the next question yeah let's take this question now hope you can all read the question clear in front of you in how many ways can 16 different books be distributed among 3 students A B and C says that B gets one more than A and C gets 2 more than B and I know that's wrong and I know you will be surprised to know that Krishna also you are wrong Anjali also you are wrong and I know you all will be surprised to know that Rippan also you are wrong I am sorry but yes it is wrong and that would surprise you also why it is wrong in fact you should tell me why you are wrong Skanda that is wrong see that's wrong so guys most of you have responded by saying okay let me solve this then I will tell you what mistake you made see when you are distributing it is assumed that everything that you have is being distributed it's not like you can retain some books with yourself okay some of distributed is also 16 by default distribution itself means you are giving away everything is that fine Arpita that basic assumption you take it and move on now listen to me all of you now here let's say A gets X number of books okay how many should we get one more than A so you will say X plus 1 how many should C get 2 more than B that is X plus 3 correct now all together these books must add up to 16 right so 2 sorry 3X I know Ravi Kiran that's wrong so 3X is equal to 12 so X is equal to 4 okay now basically what is happening over here is that you have 16 books and you are giving 4 to Mr A okay 5 to Mr B and you are giving 7 to Mr C 7 to Mr C okay now those who have done 16 factorial by 4 factorial 5 factorial 7 factorial into 3 factorial this 3 factorial is not required here you know why because now these books are not free to change hands you cannot give 4 book to B and let's say 7 book to A like that the number of books are fixed A will get exactly 4 B will get exactly 7 C will get exactly sorry B will get exactly 5 C will get exactly 7 are you getting my point you cannot change the hands of these books see this was a googly question most of you were bored out sure but why are we assuming that she is giving away all the books like shawmic distribution itself means I have to give up away all the books the word itself literally I mean from the dictionary itself means you are giving up everything that's the meaning of distribution distribution division is the same thing absolutely that's going to be the answer so this 3 factorial is not required because you are not allowing these numbers to shuffle among themselves right they are fixed they are rigid now try to recall the rationale behind multiplying with the factorial of the number of groups the rationale there was that these numbers could change positions these numbers could change groups which right now you are not allowing in this problem that's why this 3 factorial is not going to come up so your answer is just going to be 16 factorial by 4 factorial 5 factorial 7 that's it you got my point is this clear everybody please type clear now we will talk about the concept of division of distinct objects into groups where equal size groups are allowed please note that non-blank non-blank equal size groups please note that no group should be empty and there is a possibility that some groups can have equal number of items let's start with a simple example let me see how many of you are able to answer this let's say 4 people have to be divided into 2 groups of 2 to each order is not important order is not important can you tell me how many ways can I do that your response is welcome many of you are answering this with 4 C2 that's 6 let me tell you there will be not more than 3 cases let's say these 4 people are A B C D correct so you can have a group like A B C D correct you can have a group like A C B D or you can have a group like A D B C I challenge you to tell me any other division of these 4 people other than these 3 now don't tell me C D and A B because it's same as A B and C D right Krishna the answer got divided by 2 so many of you who thought that the answer would be just finding 2 people out of 4 and the other group is automatically formed this is not going to work out this is not going to work out you know why because the groups are of equal size are you getting my point so here comes a generalization of this concept all of you listen to this very very carefully so generalizing this concept if you have M N distinct objects if you have M N distinct objects and you are trying to make N groups each having M objects in them how many groups you are making N groups each with M objects in them then you answer for this when the order is not important so the number of ways in which to do that when order is when order is not important is everybody know understand the meaning of order is not important it means you are not assigning any number or any position to these groups when order is not important the number of ways of doing it is M N factorial by M factorial written N number of times and further divided by N factorial just like in the previous example you did 4 divided by 2 factorial 2 factorial and again a 2 factorial that's how it becomes 24 by 8 which is C so I am just scaling that particular concept up and here it will become M N factorial by M factorial M factorial N number of times divided by N factorial is that fine what will happen if the order is important if the order is important order is important all you have to do is take the same result that you have written above and apply it with N factorial that's it job done are you happy with these 2 formulas if yes I am going to throw some questions at you is this clear to everyone everyone over here is it clear only 3 pence is clear okay let's have a question alright here is the first question based on this concept in front of you in how many ways can a pack of cards be distributed equally among 4 players in order I am first inviting response for part A of the question time starts now I appreciate you give the answer within 1 minute Shomik come on that's a very obvious error you have made that's that's that sounds correct correct Ananya Shomik please correct your answer I am sure you can correct your answer Shisti that's not correct no Ravi so Krishna Krishna Krishna that's not correct Krishna my god Ajay is using some computer code to write star star means race to the power of 4 okay no Ajay that's not correct okay for the first part for the first part let me just tell you what happens here you have 52 cards which you are distributing equally among 4 players first of all 4 players itself means order becomes important right 4 players itself implies order becomes important implies order is important correct and since these are equal size groups your answer is going to be 52 factorial by 13 factorial 13 factorial 13 factorial 13 factorial by 4 factorial into 4 factorial okay this 4 factorial was because you have equal size groups that are 4 in number remember you have to divide it by the factorial of those many number of groups which are of equal size this 4 factorial because order is important getting my point so this 4 factorial and this 4 factorial will get cancelled off giving you the answer as 52 factorial by 13 factorial to the power of 4 absolutely shishti abhishek shanken again that 4 factorial will not come guys let's have this understanding very very clear because it will create a lot of confusion later on have I made myself clear with part A of the question if yes please move ahead and solve the part B of the question now it is correct shishti that's correct that's correct everybody's giving me the right answer awesome you guys are too good for second part your answer will be 52 factorial by 13 factorial to the power 4 divided by 4 factorial awesome I'm sure you will not have any problem doing the C part as well now I'm inviting answer for the C part yes nilag is asking me something part B same as the last part yes guys you have already started giving me the answer aditya sets guys aditya you are correct ananya no you are not correct individually I'm telling for each one of you krishna is correct shomik is correct shankin you are not correct abhishek dear why did you add 52 factorial I didn't understand that at all thrippan my dear friend why these 4 factorial and all my dear okay see guys again the logic is pretty simple I don't know whether you are able to see the pattern over here the pattern is if you are distributing it into some groups which are of equal size let's say 17, 17, 17 they were of equal size and the fourth one I was having one right so the logic is simple if the order is not important see you are not distributing it to players right nowhere in the question says you are dividing it into players we are just saying you are dividing it into 4 sets correct so here order is not important first of all order is not important so if order is not important it is just 52 factorial by 17 factorial 17 factorial, 17 factorial 1 factorial and since you have 3 groups which are having same number of objects you will further divide by 3 factorial getting my point this is just going to be your answer over done done getting my point nothing else you have to do so remember to divide by this because this was because because 3 groups had equal size getting my point getting my point everybody so please make no mistake about it this is very vital concept okay I will give you one more opportunity one more problem I will give you but before that is this clear everyone can I have a thumbs up from all of you okay ananya razer fine so next problem is coming your way let's solve this question in how many ways can 9 different books be distributed among 3 students if each receives at least 2 books if each receives at least 2 books guys be very careful when solving this question please compute your answer in exact form and tell me give me your answer in exact form exact form means I don't want in terms of NCR give me an exact form no no I never said that Anjali I never said that any books can be remaining at least everybody should get 2 books that's what I am saying please note in case of distribution you cannot retain any books with yourself you have to give away all the books that's the same thing Arpita asked a little while ago distribution means no book should remain okay you should end up giving up everything to the people no aditya that's not correct 10 no that's not correct take your time this is not a simple question looks like a 2 line question but it's actually a deep question okay I have got an answer from someone no that's not correct in fact all the cases are not covered here I think Arpita all the cases are not covered I really appreciate your effort in this but all the cases are not covered no Krishna that's also not correct no Shankin not correct not correct Rannak good guys keep trying I just like to see everybody trying no Shushant that's not correct who was that Ananya Ramachandra absolutely correct wonderful Ananya Ramachandra has given absolutely correct answer well done that's not correct so one person so far has given me the right answer that seems to be again Arpita again one very small mistake here no Abhishek, no Shusti okay now all of you please listen to this let us have cases to solve this question let's have cases to solve this question case 1 case 1 is a case where you are distributing these 9 books as 5 okay that's one case no Tripan, no Siddhartha that's not correct case 2 is a case where spitting these 9 books as and 4 correct see I am ensuring everybody is getting 2-2 books and I am also ensuring all the books actually it's needless to say distribution itself means all the books should be given away okay third case is 333 is there any other possibility I don't think so if you go and give somebody 5 books sorry 6 books then other will get 1 and 2 which will violate our condition of getting giving at least 2 books to each person okay now see here how it works first of all tell me is the order important here yes or no just type yes or no on your screen is the order important here everybody please respond yes the order important why is the order important because I am giving it to 3 students that means these books can change hands right so let's talk about the first case in first case the number of days would be 9 factorial by 2 factorial 2 factorial 5 factorial sorry my 5 became slightly turned 5 factorial further this 2 factorial because 2 groups are of equal size remember you have to divide it by the factorial of how many groups are of equal size getting my point into 3 factorial because order is important getting my point let's now go to case number 2 9 factorial by 2 factorial 3 factorial 4 factorial that's it into 3 factorial because order is important correct next is 9 factorial by 3 factorial 3 factorial 3 factorial another 3 factorial because these are equal size groups these are equal size groups into again 3 factorial because order is important now you add them whether by using your calculations or calculator I don't mind save your time hey don't use calculator please it's not allowed okay so I can't see every monitor everybody right now from here but when you add it your answer is 11508 is that clear is that clear are you happy with this solution has this question made your concept clear or not it's a very good mixture of all the sub concepts which we have taken so far under division of distinct objects under groups okay good to know that so I'm not going to begin with another concept that's going to be another important one that is called arrangement of distinct objects into groups arrangement of n distinct, distinct means different objects into are different groups okay now many of you would think isn't it what we had done just now so why is doing it again see guys couple of differences there in this concept and what we did before first of all watch out for the word arrangement we never arranged anybody in a group right right we just segregated them there was no arrangement for example a b and cd we never did b and dc and all those things correct so there was no arrangement going on in that case secondly here your objects are different that means your order is always important right thirdly in this case I will talk about two cases where blank groups are allowed blank groups are allowed and I will also talk about the next case which come later on blank groups are not allowed while in our division into groups remember first of all we were never arranging them secondly order was important or may not be important or order may or may not be important here order is important always because there are different groups and thirdly blank groups here are allowed right so we are taking both the cases blank groups are allowed and blank groups are not allowed so those who feel that this concept resembles with what we did a little while ago let me tell you there are subtle changes in this logic and the logic which we had a little while ago now instead of giving you a theory I would like to ask you as a question if blank groups are allowed and you have n distinct objects to be arranged in those are blank groups in that are different groups how many ways can you do that everybody's response is welcome blank groups are allowed yes yes I want response right or wrong guys I don't care just ok Anjali has tried no Anjali unfortunately that is not the way see first device the logic formula will always come after the logic has been devised have you ever started coding before making an algorithm no right you first make an algorithm and then you fit in your code there right ok so first collect your thoughts then convert into words right don't start using words don't speak without having a thought in your mind what is the logic here ok if you want to speak out please do so there is no restriction on the size it may see it may be of equal size it may be of unequal size I'll give you a typical example let's say there are 5 different balls ok let's say there are 3 boxes and I've named these box box 1, box 2, box 3 that means I've made the order important for them ok now how many ways can I arrange them it can be like let's say one way is this small ball and this let's say I call this ball 1 ball 2, ball 3, ball 4, ball 5 ok let's say ball 1 and ball 5 go over here ok ball 2, 3 and 4 go over here this is one way same way can be shuffled also for example now ball 5 and ball 1 can be interchanged their positions in this box 1 ok and again in the box 2 you can change the position of B2, B3, B4 also let's say B3 B2 B4 like this ok and let's say this is empty for now ok this is empty for now so these are two ways to do that so my question is how many ways can you arrange the word here is arranged you're not merely stopping by distributing these balls you're also arranging it atripan does this answer your question so there's no allocated size it may be of equal distribution it may be an unequal distribution you may end up putting all the balls in one box also and you can arrange in one box only siddhartha no siddhartha that's not correct try try see my role is not to solve questions for you let me tell you I'm not a problem solving machine my role here is to make you think how to solve the problem right because problems will change problems will change in your life so once you know how to tackle them you are independent you can fight any problem you can fight any you know hardships in your life so I'm please think don't learn how to solve one problem two problem or ten problems or hundred problems learn to think A O P S art of problem solving see if I have to solve this problem probably I'll solve the first one that will give you an idea how to go about with the second one if I have to solve this problem let me call these objects as O 1 O 2 etc till O N treat this grouping thing as if you have as if you have R minus one identical sticks R minus one identical sticks now mix these sticks with these objects mix again I'm repeating this mix these N distinct objects with R minus one sticks and arrange them in whatever possible manner you can right so first step what did I do I chose R minus one identical sticks right second step what did I did what did I do I mixed these N distinct objects with these R minus one identical sticks okay third step what did I do I permuted them okay now this is something which you already know if see if you have certain number of objects identical in a set of objects how do you find different arrangements of all the objects chosen everything at a time you take the total objects that is in this case it will be N plus R minus one objects correct take the factorial of it and divide by the factorial of the objects which are identical in shape and size right yes or no this recalls me first let me clear this off because we are going to talk about it later on this recalls me of that question that we did Mississippi it's like this only in Mississippi when you have certain number of alphabets identical how we should solve this try to recall very first class that we had on this we counted the total alphabets here one two three four five six seven eight nine ten eleven and we should divide it by the factorial of the number of alphabets which were repeated so four i's are repeated so four factorial four s is repeated so four factorial two p's are repeated two factorial do recall this concept do recall this the same thing I am using over here also you have N different alphabets think like that and R minus one identical alphabets how many different words can you make from them using all the alphabets together you will say N plus R minus one factorial by R minus one factorial getting my point trippin right so you have to further divide your answer with R minus one factorial because R minus one are identical objects now you would ask me sir how does it take care of blank groups it's very simple if let's say two sticks come together anywhere let's say I am just taking one of the situations o one o two and you had one two three six coming over here then o three o four o five then let's say one stick coming over here o six then again two sticks coming over here let's say o ten like that let me not write it in order because you are free to write any objects in any position so let me just make it as o o seven and o let's say ten sorry eleven like that dot dot dot dot now here if you see the first group this is your first group it has got two objects second group is blank blank means empty no object third here is blank fourth here has three objects right fifth this is your fifth group fifth has got o six sixth is again blank like that so basically moment two sticks come next to each other there is a blank group created in between are you getting my point everybody is convinced with this result so please make a note of this please make a note of this the number of phase you can arrange it is n plus r minus one factorial by r minus one factorial so here you are taking care of their arrangement also here you are taking care of the fact that they are divided into different groups also and here you are taking care of the fact that blank groups are allowed also guys make sense happy or not clear okay if you're clear with this now I leave up to you to tell me how many ways can you arrange and distinct objects into our different groups where blank groups blank groups are not allowed now I don't want blank groups either you can write it on your notebook and send me a snapshot or you can just type it on your chat box or you can unmute yourself and speak out so many modes of communicating with me okay no shawming that's not correct minus one c very slight error is there in your answer you are very close to the answer just one small mistake you have done why do we do that shankin because there would be no blank groups you would eliminate all the number of blank groups just means shankin that you should not allow any two sticks to come together and you cannot have sticks any stick before the first object and you cannot have any stick after the last object now that's correct absolutely correct that's correct so trippin has already given the right answer no angeli that's not correct shawmic no that's not correct okay let's take this up see here now just now I was explaining shankin that if you want to have no blank groups then no two sticks should come next to each other right that means if let's say your objects are o one o two o three o four da da da da da til o n the sticks can only come in the positions which are marked by these crosses okay let me just write an object before this also who will tell me how many crosses have I written that means how many gaps are there n minus n minus one gaps are there right n minus one gaps are there now how many sticks you have we have r minus one sticks correct to create our groups we need r minus one sticks right you have to choose any r minus one gaps from these n minus one gaps right so we need to choose we need to choose r minus one gaps from these n minus one gaps to place r minus one sticks right and these objects can be inter arranged so o one o two etc. till o n can be arranged among themselves in n factorial in n factorial ways okay so just combine these two so the number of ways in which you can arrange r distinct objects in r different groups such that brand groups is not allowed will be n minus one cr minus one into n factorial absolutely correct shawmic now this is the answer well done trippan was the first person to give this answer correctly yes here everyone can we have a quick question on this let's have a question on this yes there goes the question in front of you in how many ways can five different balls be arranged into three different boxes so that no box remains empty so that no box remains empty sorry getting the response shawmic is absolutely correct siddhartha is also correct krishna krishna krishna which is your answer krishna yeah second one is correct correct shankin correct trippan almost everybody who has replied is correct shashank why did you divide by two factorial here see it's just the use of the formula see no brand groups are allowed over here so that is n factorial n minus one cr minus one right here n are your balls which is five in number r are your boxes which are three in number so your answer will be five factorial by four c2 that's 120 into six that's 720 that's the answer are you getting this point okay so guys before we move on to the next concept a good news for you I'm giving you a five minute break so please have some water or cup of coffee whatever you want five minutes break exactly 40 will resume will resume at 5 40 p.m so when will class end see will end the class at 6 30 for non KVPY people and 7 30 for the KVPY people okay sir alright welcome back break is over guys hope you are back back with your notebook and pen in front of you screen in front of you so now we are going to talk about again a very similar sounding concept the similar sounding concept now is how do you distribute how do you distribute and distinct objects and distinct objects into our different groups our different groups okay now when I say distribute it is just segregation there is no arrangement happening correct again this may sound very similar to the concept of division of objects into groups but remember in division of objects there is no groups which are blank but here we are going to allow blank groups okay so first thing is I'm going to talk about distribution of n distinct objects where blank groups are allowed blank groups are allowed okay and we will also talk about where blank groups are not allowed secondly how it is different from the concept that we learnt in the beginning of this class today where we are talking about division of people under different different groups is that there the size was known right there you could know the size of the groups but here there is a limit to the size that means any group can be empty also any group can contain any number of items also correct so this the size of the groups here are first of all equal unequal that is there and also we are not aware of the size are you getting my point okay so let me pose this as a question to you all if you want to distribute n distinct objects into r different groups with blank groups allowed how many ways can you do that it's a very simple concept guys I'm sure you'll be able to answer this within 10 seconds if you think it in the right direction so think as if there are okay groups like this group number one group number two da da da da till group number r okay so this is your arith group okay and there are n distinct let's say objects o1 o2 o3 etc till on which is supposed to be distributed among these groups distributed remember the word is distributed no arrangement only distribution says that blank groups are allowed and there is no information about the size of these groups that means you don't know how many objects are going to get into group one or how many objects are going to get into group two etc can we have the answer for this I'm inviting everybody to write the answer tell me the answer for this time starts now I think one minute is max I can give you because it's very simple concept yes anybody please feel free to answer answer me privately but I want to see your response let it be wrong now remember here the constraints blank groups are allowed secondly you don't know what is the size of each group you don't know which objects are going to get into which group so how many total ways to distribute n distinct objects into r groups guys get this distinction very clear when I was preparing for je when I came across these concepts they all sounded very similar to me but no they are not okay so I have a response n c n minus r into no that's not correct no trippin that's not required guys the answer to this will shock you it's super simple every object over here has n r choices right if you talk about object one how many choices does object one have it can go to group one or group two or group three till any of the r groups it can choose correct so this object o one has r options that means it can get into any one of these r groups correct now object two also has r options object three also has r options because see let's say object one gets into group number three object two also can get into same group also correct so it doesn't mean that since object one has got into group number three it will stop object two from coming inside that group correct so every object of these objects have r r options each that means the total number of ways would be nothing but r into r into r n number of times that is r to the power n ways absolutely correct siddhartha super simple right I knew many people would you know take it to a different level and start thinking at a different trajectory altogether correct okay now what will happen let me take the case that is the second case where blank groups groups are not allowed blank groups are not allowed anybody has any idea about how to do this any idea this is a concept which is not very easy siddhartha I appreciate your answer but that's not correct okay now all of you please listen to me okay this is a slightly challenging concept that's why I am putting three stars to it okay all of you listen to this concept in order to solve this question we have to use a very important principle that we call as the principle of inclusion and exclusion okay now listen to this concept because it's going to be used again this is going to be used again in derangements concept which is coming later on for you principle of inclusion and exclusion okay this is very important concept okay this concept is in short form called pi principle of inclusion and exclusion now let me explain this concept very slowly and please listen to me very carefully okay all of you please listen to me very very carefully if blank groups are not allowed this is what I am trying to solve okay so let's say we have we have these are groups with us okay we have these are groups with us these are your groups okay first of all let me ask you this question if there is no restriction at all if there is no restriction at all how many ways can you distribute these n objects into these r groups we just not figured it out it was r to the power n correct so this is the number of ways when there is no restriction at all okay let me call this as the universal set okay now I am talking in terms of sets because I am soon going to use a very important formula for you okay so this is when there is no restriction there is no restriction at all okay now let me ask you this question let I be the event or let's say let G1 be the event when group 1 is empty group 1 is empty or blank okay then tell me in how many ways can you distribute the rest how many ways can you distribute the n items ensuring that the group 1 is definitely blank now remember when I am saying group 1 is definitely blank the other groups may or may not be blank right Trippan Trippan says r-1 to the power of n do you all agree with him I think he is correct right because all the items or all the objects have full liberty to go to other groups but definitely not group number 1 so group number 1 is definitely blank okay similarly if I ask you what is the what is the number of ways in which group 2 will be blank definitely blank now others may or may not be blank remember it will be same r-1 to the power n it will be true for all the groups let's say gr group correct now I am going to ask you next question please answer this very carefully let g1 intersection g2 be the event that g1 and g2 group are blank can you tell me in how many ways can this happen where you definitely know that g1 and g2 are blank others may or may not be blank what will be the answer for this r-2 to the power n correct absolutely correct same will be true even if there are let's say g1 and g3 also r-2 to the power n so there is no doubt about this whatever combination of 2 you take let me generalize this gi intersection gj the answer will be r-2 to the power n okay if I again scale this concept and say let's say g1 intersection g2 intersection g3 is the event where groups g1 and g2 and g3 are definitely blank others may or may not be blank then how many ways can you distribute your objects r-3 to the power right absolutely you will say r-3 sorry r-3 not n-3 r-3 to the power n let's say any 3 pairs you take i, j, k this will be again r-3 to the power n this process will continue on I am not going to write this on and on you must be thinking what am I going to do next what I am going to do with all these things that I am writing right now on the screen see here what is your requirement your requirement is you don't want any blank group that means you want where g1 doesn't happen and g2 doesn't happen see what was g1 event g1 event was your fact that g1 is blank now you don't want your g1 to be blank so g1 compliment I am just talking in terms of language of sets guys I am not talking about I am not talking rocket sense and you don't want g2 to be blank and you don't want g3 to be blank and so on and you don't want your gr to be blank right this is what we need we need this ok now according to de morgan's law can I say it is n of g1 union g2 union g3 all the way till gr whole compliment yes or no yes or no compliment means total number of events minus n g1 union g2 union g3 etc till gr yes or no now do you recall a formula that you had studied in sets the formula was this formula I am just writing it on top over here I am sure all of you had done this formula n a union b union c right what was this formula this formula was n a n b plus n c minus n a intersection b n b intersection c n c intersection a plus n a intersection b intersection c do you recall this formula I am sure this formula is extendable to any number of sets you want you just have to do let's say there are more sets like a1 a2 a3 till let's say ar all you have to do is add them one at a time that is you just do summation of ai then subtract summation of intersection of 2 at a time that is a1 intersection a2 then add summation of intersection of 3 at a time a1 a2 a3 keep doing this this is what is actually the inclusion exclusion principle you include then you exclude then you exclude like that so I can say that but in light of that formula this can be written as n union summation of n g1 minus summation n g1 intersection g2 plus summation n g1 intersection g2 intersection minus dot dot dot I am not going to write all the terms hope you have understood the rationale behind it now what is nu nu means no restriction no restriction we just showed this one is the no restriction this is nu this is your nu so this is going to be let me scroll down this is going to be r to the power n minus now summation n g1 as you can see all your gis are r minus 1 to the power n so how many such groups are there r groups are there so you can do r into or rc1 into r minus 1 to the power n correct minus 2 at a time is nothing but r minus 2 to the power n but how many 2 groups can you choose from r groups rc2 ways if you see the trend it goes like this and of course it will stop somewhere where your r will start becoming where this term will start becoming 0 are you getting my point okay so this is the formula for finding the number of ways of distribution of distinct objects into r different groups where bland groups are not allowed this formula i would request all of you to please make a note of getting my point everyone is this theory clear to all of you please type clear i would love to see if it is clear to everyone here okay this formula is going to be used next year also for you when we are doing the functions chapter okay let's solve questions to understand this concept better let's have this question let's solve this question wonderful question in how many ways can 5 different books be tied in 3 bundles looks like a very simple question correct but it incorporates the same concept which we have just now dealt with of course they have to have books in dharta how can you make a bundle which doesn't have a book in it but by the way that's a good question to ask so yes the answer to this is all the bundles need to have books in them guys please give your answer in the exact form don't give me a long expression i'm sure you can evaluate it the answer is pretty small so everybody is ah sharmic that's not correct and you would be shocked to know that sharmic that's not correct you have made a very fundamental error there read the question the question is 5 different books can be tied into 3 bundles ah no siddhartha you have done the same mistake as what sharmic has done trippan also does the same mistake shankin also did the same mistake ananya has given a different answer but that's wrong but one thing i like about all of you people are consistently there in responding no rabhi that's wrong no vibhav that's wrong that's also wrong skanda skanda now i'm sure you are shocked to know that the answer that you gave was wrong okay i'm sorry to say but yes it was wrong i'll tell you the mistake also see it no trippan that's also wrong i'm sure you tried dividing my three right okay see here first of all when you say you are dividing 5 different books into 3 bundles you actually mistook the concept as distributing 5 different objects into 3 different groups but here the different groups means they have an order now these bundles there is no order to these bundles we are not calling it as bundle 1 bundle 2 bundle 3 now you are correct trippan now you are correct are you getting it i purposely gave this question because i knew everybody would be tricked by the concept if you are distributing 5 different objects into 3 different groups see the name of the topic which i gave i'll go back sorry i'm going back to the previous world just i wanted to show you the name of the topic the name of the topic is okay i have to go back one more page board number 16 yes see here categorically mentioned are different groups different groups means order matters order is important no shankan that's wrong shomic that's also wrong okay now see let me complete it guys so that you get an idea about where you went wrong that's correct siddhartha so when you did this formula r to the power n minus rc1 r minus 1 to the power n minus rc2 r minus 2 to the power n you ended up getting this answer right by the way here n is 5 r is 3 so you have got 3 to the power 5 minus 3 c1 2 to the power 5 then minus 3 c2 1 to the power 5 next term will not even appear okay so forget about it okay so 2 to the power 5 will be 243 this is going to be 32 times 3 which is 96 and this is going to be minus of 3 plus sorry plus okay so this is 246 minus 96 that's going to be 150 but this answer is not complete because you have considered those bundles to be different bundles correct so your answer would be 150 divided by 3 factorial because you considered those 3 bundles to be different while you are getting this answer 150 so you have to compensate by dividing by 3 factorial that means your answer will be 25 not 150 I am getting this point you understood now where you went wrong everybody so it is not into 3 it is not divided by 3 it is divided by 3 factorial because now these bundles even if they switch the books among these there are no order assigned to those bundles there is nothing called bundle 1 bundle 2 and bundle 3 you are just segregating it to 3 bundles so this is not different groups there are groups but there is no order to it getting my point let's have another one now before you solve this problem a bit of theory I would like to discuss with you because I am sure many of you would not understand all of a sudden what is this meaning on to what is this word on to all of a sudden just a brief explanation before you solve this I am sure you will be able to solve it first of all mapping means functions correct so how many functions can you form which maps elements of set A to set B says that these functions are on to in nature on to just means just a plain and simple explanation for on to functions on to function means co-domain is equal to your range that's it that's it all you need to know is on to means co-domain is equal to the range that means all the elements of your set B let me since there are 3 elements let me call it as B1, B2, B3 there are 5 elements here let me call it as A1, A2, A3, A4, A5 that means all your B1, B2, B3 should become your range now please attempt to solve this okay 75 mhmm trippan you are correct shankin no siddhar ka no absolutely correct skandha that's correct no ananya that's not right see 3 to the power 5 would have been the answer when your range can be anything that means all of them can map to B1 also okay here you have to maintain the fact that all B1, B2, B3 should be used now treat as if these are balls and these are groups okay so end distinct balls and these are 3 different groups correct by the way n is 5 over here so let me write 5 now these 5 different balls have to be placed in these 3 different groups such that all the groups are used up that is the meaning of the fact that core domain is equal to range right isn't it a plain and simple concept of using our formula of distribution of end distinct objects into our different groups where band groups are not allowed that formula is again 3 to the power 5 minus 3C1 2 to the power 5 minus 3C2 1 to the power 5 etc that's again 150 so I am not going to solve this now let's say there is a person who doesn't want to use this formula okay he says sir anyways I am going to forget this formula in the exam so can I solve this by logic just by looking at this question no formula okay I will solve this question by logic here you have to make cases assume that these b1 b2 b3 are like your 3 boxes okay where you have to place these 5 objects such that no box is empty so I have following cases case number 1 I put 1 ball in this 2 ball in this correct case number 2 I put 1 ball in this okay 1 ball in this and 3 ball in this okay and I say order is important that means I have allowed their shuffling also correct can I now solve this question by the very first concept that I discussed today that was second concept which I discussed today that was division of objects into groups can I use that concept to solve this of course we can and I am going to see solve this concept by using division of objects into groups and get the same answer see how am I going to do this so think as if there are 5 objects and you have to distribute it into 1 2 2 and order is important so how will you solve this you will say sir 5 factorial by 1 factorial 2 factorial 2 factorial addition by 2 factorial because now they are 2 identical size groups right into 3 factorial because order is important correct that's the solution for case number 1 right case number 2 case number 2 who will tell me 5 factorial by 1 factorial again 1 factorial 3 factorial again 2 factorial because now you are distributing it like this let me draw a distribution diagram 1 1 3 and these are equal size groups these are equal size groups that's why divide by 2 factorial into 3 factorial because order is important correct let us calculate these so this is 120 by 8 into 6 that's going to be 15 into 6 which is 90 this is 3 factorial 3 factorial gone 120 by 2 which is 60 cases means addition added your answer is 150 same way you got the answer same answer you got now by using the concept of division right understood yes sir clear I would really appreciate if you type clear so that I know how many of you have taken this concept well excellent now guys in the next 15 minutes of your class I am going to pitch in another important concept after that we will end the session of PNC for everyone and for KVPY people we will start with the concept of pair of straight lines fine so the next concept which we are going to talk about is distribution of see all these concepts they will sound very similar to you but there is a difference between these concepts distribution of identical objects into let's say n identical objects into are different groups or you can say are different people so that it sounds the same thing our different groups is our different people okay group or people the moment you say people it means group with order being important people means group with order being important order becomes important okay okay let me ask you as a very simple question I have 10 chocolates all these chocolates are identical so let's say I have 10 ferro rocher chocolates 10 ferro rocher chocolates now there are 3 needy children who need these chocolates let's say I call that those children as XYZ why am I writing plus so these are 3 needy children okay they make them different so 1 here 2 here and 3 here okay you have to distribute these 10 ferro rocher chocolates which are all identical needless to say that among these 3 XYZ children how many ways can you do that logic first logic anybody can unmute yourself and speak out say this is the logic that I have in my mind to solve this you must be asking are blank groups allowed here can they be a person without a chocolate also yes let's say blank groups are allowed blank groups are allowed that means there can be a person without a chocolate also okay so let's say if I feel like I give all the chocolates to X and I give no chocolates to Y and Z that can be one case sir we can do that we can do this thing and permute all the sticks permute do you need to no combine okay even if you permute I think it was Trippan right yes sir so Trippan has an idea Trippan says let us have since we have 3 children I need 2 sticks let us try to place these sticks between these so there are n balls or identical chocolates let's say 10 identical chocolates my bad and let us have 2 identical sticks 2 identical sticks okay okay now just mix them up and permute them in as many possible ways as you can okay remember if you have 10 plus 2 12 objects out of which 10 are identical of type 1 you have to divide by 10 factorial and 2 are identical of type 2 then you have to divide by 2 factorial so your answer is nothing but 12 C 10 or 12 C 2 whatever you want to call it that's nothing but 66 correct is this clear is this part clear in your mind then only I will be able to scale this up to n objects and r groups now you must be thinking how do I account for the fact that there can be a student or there can be a child who doesn't get anything see when 2 sticks come together basically between that 2 sticks and if those 2 sticks come just in front then let's say I put these sticks like this stick stick and all the 10 chocolates it means the first student X or the first child X doesn't get anything 0 chocolates for him second also doesn't get anything 0 chocolates Z guy has got all the 10 chocolates with him understood if you have understood you should be able to scale this concept and tell me in how many ways can I can I distribute n identical objects into r different people or r different group such that blank groups are allowed such that blank groups are allowed absolutely so n plus r minus 1 factorial by by n factorial r minus 1 factorial also you can write it as n plus r minus 1 cn or n plus r minus 1 c r minus 1 both are correct so please remember this expression very very useful everybody is convinced today we are learning something which is all sounding very similar to each other okay now in the remaining 5 minutes of our class sorry 7 minutes of our class can we have answer for the fact where blank groups are not allowed that means there should not be any child who doesn't get a chocolate sorry please type in your response or you can speak out no don't speak out if you speak out everybody will get to know the answer just type it down or send me a photo on whatsapp does it take so much of time I don't think so n minus 1 n minus 2 da da da da no injury that's not correct tripping you are absolutely correct bang on good fast fast fast I don't think so this is difficult so far just tripping has replied with the right answer siddhartha no I think you just thought it in a very complicated way same as shawmik also you have thought it in a very complicated way shawmik see here if you don't want any blank groups let's say these are my 10 chocolates okay of course these 10 chocolates let's say I start with that 10 chocolate example okay now if you see the gaps between them you can only place your 2 sticks in these 9 gaps okay there are 9 gaps there are 9 gaps and the moment you choose any 2 of these 9 gaps your distribution is done completed getting my point let's say I put my stick over here and let's say I put my stick over here the moment I choose these 2 gaps automatic segregation of these chocolates 3 people where the first guy gets 1 chocolate the second guy gets 3 chocolate and the rest 6 chocolate goes to the third guy correct now there is no inter arrangement possible because the objects themselves are distinct they are identical right it's a very simple concept you just pick up 9 gap 2 gaps from these 9 gaps job is done so when you have n objects when you have n objects and you have r-1 sticks you just have to pick up r-1 gaps from n-1 gaps and your job is done this is your answer this becomes your answer so these 2 formulas please please remember this very very helpful is that clear so those who are thinking in a very complicated way it's good that you are applying your brain but never forget obvious things right my teacher always used to say that solving a problem is like catching a ball if you run faster than the ball or run slower than the ball in both the cases you don't catch the ball probably most of you ran faster than it because I can see lot of concepts being applied let me just seal in this session with a simple problem very simple question we will take up and then we will call it the end of the class for non KVPY people of course for KVPY people we have more just a simple question because there is just 2 minutes left I am sure you should be able to do this 4 boys picked up 30 mangoes in how many ways can they divide them if all mangoes are identical just tell me the expression in terms of NCR triple that's not correct shawmic is correct ananya not correct mayur is correct krishna is correct skanda not correct shankin has actually calculated it no shankin that is also wrong so far krishna mayur shawmic that's correct guys come on all you need to do is use that formula N plus R minus 1C R minus 1 or N plus R minus 1C N N is the number of objects to be distributed mangoes R is the number of boys so 30 plus so that they are dividing them so no blank groups are left dividing them doesn't mean just means you have to give up all the 30 mangoes it doesn't mean all the groups have to be filled up yes so the answer is 33 C3 or 33 C30 both are correct okay fine guys so for the non KVPY people we'll end the class right here I'll share this PDF with you after 730 when the class is completely over okay next class stay tuned for more concepts that are coming up and in case you start your school I'm sure you'll not start because Friday is your last exam right okay anyways after October 8th only you'll come to know what is the new chapter starting anyways we are ahead of the school okay thank you those people who are going to leave this session thank you sir bye bye just 5 minutes break for the others before I start KVPY 635 I'll resume okay thank you so much bye yes somebody was asking me something okay so people who have stayed back for KVPY session let us resume pair of state lines I think last class I introduced to you what is pair of state lines okay also the concepts which are going to take in this class are very important the first concept that I'm going to talk about is the concept of angle between angle between the lines which constitute those pair of state lines let me write it as angle between the pair of lines for that skanda you need to watch the video because we had one session on this topic what I will suggest to you is stay in this class no worries but I'm not sure whether you will understand lot of things so just after today's class or whenever you get time as early as possible just watch the previous class last one our lecture okay the link is there in the group sir can you just tell what we ended quadratic with quadratic we ended quadratic with the concept of location of roots okay so let's say we have these two lines let me just draw these two lines like this and like this okay now we want to find out the angle between these two lines okay now first of all how many of you I did I tell this last class that if a pair of state lines let me write it like this if a pair of state lines state lines inter-6 at origin okay that means this point is origin then out of these terms that we normally write for a general second equation for any conic which also includes pair of state lines let me be clear with you that when your pair of state lines is passing through origin that means they are intersecting at origin you will not have these three terms appearing it will only have x square x y n y square term in it correct so we normally call a pair of state lines intersecting at origin as a second degree homogeneous equation as a second degree homogeneous equation please note homogeneous means the degree is uniformly throughout so as you can see a x square 2 h x y b y square all of these are second degree terms correct there is no first degree term like 2 g x 2 f y and there is no zero degree term like a concept they have all been removed now can somebody tell me why did I remove these three terms now for that you need to understand a very simple concept that if your line was passing through origin then both of your lines would be of the nature y is equal to m 1 x and y is equal to m 2 x right we all know that lines passing through origin they don't have a constant term in it right so let's say slope of this line is m 1 slope of this line is m 2 so one will be y equal to m 1 x other will be y equal to m 2 x correct if you want to form a pair of state lines from this I already told you that when you have two constituent lines you want to form a pair of straight lines from it you just multiply it by converting both of them into a general form that means take everything to the left hand side okay if you multiply it you will end up seeing terms like y square you will end up seeing terms like m 1 m 2 x square and you will end up seeing terms like let me write it at a slightly distant apart y square and you will see terms like minus m 1 plus m 2 x y also correct as you can see it doesn't have any linear term like 2 g x 2 f y it doesn't have any constant term like c so this is actually a homogeneous second degree equation getting my point that's why I dropped off 2 g x 2 f y and constant term from that okay now you must be wondering hello sir has done this how does it help us to find the angle between them guys it's obvious now everything is in front of you you know the slopes of these two lines the slope of these two lines is m 1 m 2 correct what is the angle between two lines whose slope is m 1 m 2 what is the formula that we have learned tan theta is mod m 1 minus m 2 by 1 m 1 m 2 yes or no now you must be wondering how do I know m 1 and m 2 right because all we know is a to h and b because your equation will be given to you in this form right they will never tell you what is m 1 and m 2 right but never mind we can find these terms very easily how let's see can I say from this equation let me put a smiley on top of it can I say that from this smiley equation compare the coefficients if you compare the coefficients then can I say compare this with this so let me name it equation 1 and the smiley equations equation 2 so compare compare 1 and the smiley equation okay a by m 1 m 2 will be equal to 2 h by minus m 1 plus m 2 will be equal to b by 1 yes or no what I did I took the ratio of these coefficients guys never do a is equal to m 1 m 2 correct never do 2 h is equal to minus m 1 the equations are equal that doesn't mean their coefficients are equal their coefficients are proportional okay very simple thing but many people I see making mistakes in it okay now can I say from this equation my m 1 m 2 is a by b right and my m 1 plus m 2 is minus 2 h by b correct these two will actually help me to evaluate this correct let's see how will I able to do that so tan of theta first of all can I say mod m 1 minus m 2 okay and mod 1 plus m 1 m 2 correct is there any question okay now I can say mod m 1 minus m 2 square is m 1 plus m 2 square minus 4 m 1 m 2 m 1 plus m 2 square is nothing but minus 2 h by b square minus 4 a by b correct this is m 1 minus m 2 square so can I say m 1 minus m 2 mod will be under root of this thing which is 4 h square by b minus 4 a by b okay I've actually reached the end of the page so I'll write this again properly so it is under root of 4 h square by b minus 4 a by b is that fine yes is that fine clear okay any questions anyone sir it will be shouldn't it be 4 h square by b square once again 4 h square by oh I'm so sorry so sorry yes correct so I'm replacing that over here under root of 4 h square by b square minus 4 a by b okay I don't need a mod over here because it's only positive by a 1 plus m 1 m 2 is a by b getting the point okay now this is further subject to simplification first of all 4 I can take out multiply throughout with b square on the top so it becomes h square minus a b and come out as a b which will get cancelled with the b in the denominator please note this result down it's a very important result when you know the equation of the pair of state lines as a x square plus 2 h x y plus b y square equal to 0 and you have been asked to find the angle between the two lines the angle between them is given by tan theta is equal to please note this will always give you the acute angle needless to say that will be 2 under root of h square minus a b by mod a plus b is that fine everyone please make a note of this formula very important formula is that fine now an obvious question would have arisen in your mind that had that line not passing through origin let's say this point is not origin that means my equation of these pair of state lines is not a homogeneous second degree term but it also has a g x f y and a constant term in it then how do we find the angle between them any idea anybody can unmute yourself and speak out so tripan is there, ajya is there, ananya anjali, kiritana, pratham shomik, siddharta, skand srishti, shushantes shushantes and vibhav and tripan how would the result change or will it actually change yes guys I want to know your opinion any idea will it actually matter see very simple let's say instead of y equal to m1x had this line be y equal to m1x plus let's say c1 and this line would have been y equal to m2x plus c2 correct and let's say this line was actually made by the multiplication of these two equations correct, yes or no if you multiply everything you would realize that these first three terms would actually remain the same y square okay whatever we had written in the previous slide y square or you can write m1, m2, x square first since we have x square in the first term m1, m2, x square xy term will be minus m1 plus m2, xy right third term would be y square and the rest terms which is like contributing towards 2gx, 2f, y and c would be made up of c and m1, m2 a combination right so there will be other terms like x y and some constant type of thing okay now when you find the angle between these two lines does the formula change no the formula doesn't change it still remains m1 minus m2 by 1 plus m1, m2 correct and these values are going to come again from these three terms only so these three terms are important to us these terms are not important to us because they will anyway not contribute towards getting my m1 m1 minus m2 and 1 plus m1, m2 so the idea is the formula that we had derived that is under root 2 under root x square minus a b by mod plus a plus b will remain the same same formula will be valid for even in this case the other way to understand is if you shift your origin to some other point let's say x1, y1 point it doesn't change the angle between the lines does it no right so don't get perturbed even if your equation given to you is not the equation of a homogeneous second degree that means your lines are not passing intersecting each other at origin even then the formula of the angle between the lines do not change are you getting this point okay now I have two conclusions based on this formula first of all clear please type clear if it is clear so let me make two important conclusions for me conclusion number one if h square is equal to a b what does it mean that means the lines are parallel or coincident getting my point that means either they lie on top of each other or they are parallel to each other now many people ask me sir parallel means they will not intersect each other right then how can a pair of straight lines be formed please let me tell you intersection of the pair of straight lines is not important two parallel lines equations if you multiply you still get a pair of straight lines okay so don't be under the impression that your lines have to intersect even these two lines like this will constitute a pair of straight lines so multiply their equations and get a pair of straight lines equation okay second conclusion that I want to draw is if a plus b is 0 which implies the lines are perpendicular to each other the lines are perpendicular to each other obvious reason because in the formula tan theta is equal to 2 under root of h square minus a b by a plus b if your denominator vanishes theta becomes 90 degree here theta becomes 90 degree here are you getting this point is that clear okay now let us look at questions based on this first question first question I just write it down some important things on the board I will dictate it and just write down the important terminologies if the sum of the slopes of the line if the sum of the slopes of the line given by x square minus 2 c x y minus minus 7 y square is equal to 0 is 4 times their product is 4 times their product okay fine c if the sum of the slopes of the lines given by this pair of straight lines is 4 times their product fine c please answer privately Ananya you are right Ananya you have just made a sign mistake check that is it only two people have responded I need one more response 98 no idea that is not correct it is a single digit answer 2 that is correct Anjali that is correct see it is very simple question I have already given you all the formulas over here sum of the slopes is minus 2 h by b product of the slopes is a by b correct a here is 1 2 h is minus 2 c and b is minus 7 all you need to do is plug in over here correct so it is given that the sum of the slopes is 4 times their product right that is minus 2 h by b that is 2 c by minus 7 is equal to 4 times a by b a by b is 1 by minus 7 minus 7 minus 7 goes c is equal to 2 that is it done that is it clear next question if the pair of straight lines a x square plus 2 h x y plus b y square equal to 0 is rotated about the origin is rotated about the origin through 90 degrees through 90 degrees okay find the equation of the new pair formed that is correct trippan no it does not matter whether you rotate clockwise or anticlockwise does not matter there is no b in your answer okay let me solve this anybody else apart from trippan who is willing to respond no no I am not talking about these lines I am saying let's say earlier the line was like this okay now I rotated it in such a way that I rotated it 90 degrees in this way so this this line became like this let's say like this and this line let's say became like this so the yellow lines are basically the 90 degree rotated versions of the old line so this is your 90 degree rotation this is your 90 degree rotation this guy this let's say called L1 L1 became your L1- L2 became your L2- so I want the combined equation of L1 and L1- and L2- now see here again the idea originates from the fact that let's say the line represented by L1 and L2 have slope of M1 and M2 okay everybody has responded then I can clearly say M2 is minus 2h by B and M1 into M2 is A by B correct now I want that first of all this origin remains the same I want that line equation which is 90 degree to L1 and 90 degree to L2 so if this is y equal to M1x and this is y equal to M2x would you all appreciate that this line would be this line would be y is equal to minus 1 by M2x and this line would be y is equal to minus 1 by M1x correct yes or no correct so if I want to formulate a equation which is pair of straight lines for this I'll just have to multiply these two equations if you multiply it becomes y squared it becomes x squared M1 M2 and it becomes xy 1 by M1 plus 1 by M2 okay but M1 M2 I cannot leave it like this I have to put it in the values known to me this and this so it becomes y squared plus x squared M1 M2 divided means divided by A by B so B will go up okay xy this will become M1 plus M2 M1 plus M2 is minus 2h by B by M1 M2 is A by B let's simplify this further Bx squared by A squared and this is nothing but minus 2h xy by A is equal to 0 multiply throughout with A squared so A squared y squared plus B squared x squared minus 2h xy is equal to 0 is your new equation which is nothing but if you simplify it further it is if you rearrange it further not simplify it further this is what you are going to see there was an A here there was no A squared there was A only so A y squared is equal to 0 this is your answer is that fine sir we can also use the rotation of coordinate yeah you can always use the rotation of coordinate correct next find the acute angle between the lines whose combined equation is find the acute angle between the lines whose combined equation is 2x squared plus 5xy plus 3y squared plus 6x plus 7y plus 4 is equal to 0 tan inverse of the answer that they have given is quite simple tan inverse some fraction they have given simple fraction that's correct Ananya that's correct anybody else ah no Shovik okay guys this is a very simple answer tan inverse 1 by 5 see here first of all A is 2 H is 5 by 2 B is 3 okay and angle is given by 2 under root of h squared minus AB by mod A plus B okay that's 2 under root h squared is 25 by 4 and angle is 6 by A plus B is mod 5 that's already positive so we need to write mod there so this becomes ah if you take the LCM it becomes 1 by 4 so 2 into is 1 1 by 5 so theta is going to be tan inverse of 1 by 5 clear easy one okay now our next question for you all if let me ask you this question prove that the equation of the pair of bisectors for the pair of straight lines A x square plus 2 h x y plus B y square equal to 0 is x square minus y square by A minus B is equal to x y by h now let me explain you this question first so let's say we have these 2 lines okay whose combined equation is given to you over here okay now you create a bisector of the angle between these 2 lines so let's say I bisect this angle and I bisect this angle so I get these 2 bisector lines let me call them as B1 B2 lines now prove that the combined equation this is the combined equation of B1 B2 lines time starts now one once you are done so that I can discuss it any idea anybody making any progress any success see we have already learned the concept of finding the equation of bisectors right so let's say I call these 2 lines y equal to m1 x and y equal to m2 x okay it's very obvious that these lines are passing through origin because it's a second degree homogeneous equation correct now if somebody says tell me the equation of the bisectors of this line we already know that it is y minus m1 x by under root of 1 plus m1 square is equal to mod of y minus m2 x y under root of 1 plus m2 square correct all we need to do is write this in a very simplified fashion and remove m1 m2 right so let's do that okay first of all let me remove this mod thing if I remove this mod thing I have to compensate by putting a plus minus sign in front of it okay right now this itself gives you 2 equations so one equation will be y minus m1 x by under root of 1 plus m1 square plus y minus m2 x by under root of 1 plus m2 square equal to 0 and other root would be other equation will be y minus m1 x by under root of 1 plus m1 square minus y minus m2 x by under root of 1 plus m2 square okay equal to 0 if you want to create a pair of straight lines on them we just need to multiply these 2 correct multiply it multiply okay let me remove this cross else people they think it's a wrong thing okay so when you multiply it sure when you multiply you can see your a minus b a plus b formula featuring in over here okay so that's going to be y minus m1 x by under root of 1 plus m1 square minus y minus m2 x by under root of 1 plus m2 square equal to 0 is that fine let's simplify this further okay at the back of this mind please keep that m1 plus m2 is minus 2h by b and m1 m2 is a by b keep this at the back of your mind okay so when you open this up let me just cross multiply these terms let me take the LCM and all so 1 plus m2 square y minus m1 x square minus 1 plus m2 square m1 square y minus m2 square equal to 0 okay if you take your y terms y square terms you'll end up getting m2 square minus m1 square y square if you take your xy terms okay from here I will get minus 2m1 1 plus m2 square and this is plus 2m2 1 plus m1 square sir it should not be 1 plus m1 whole square I'm sorry sir it shouldn't be 1 plus m1 whole square in the first one but 1 1 gets cancelled no 1 1 got cancelled no that's why I removed okay next term would be x square this would be m1 square 1 plus m2 square minus m2 square 1 plus m1 square correct okay let us simplify this let us simplify this so m2 square minus m1 square y square first of all I will write this as m1 minus m2 times m1 plus m2 okay take minus 2xy common and open the bracket so it becomes m1 minus m2 plus m1 m2 m2 minus m1 and here it will be x square again m1 square minus m2 square which is m1 minus m2 m1 plus m2 okay can we drop the factor of m2 minus m1 from everywhere so let me drop m2 minus m1 from everywhere m2 minus m1 gone m2 minus m1 gone from here and this will give you a minus 1 this will give you a minus 1 so I am writing a minus 1 here m2 minus m1 gone from here and this will give you a minus sign over here okay now let us put in the values of m1 plus m2 and m1 m2 so I am writing it over here m1 plus m2 is minus 2h by b y square minus 2xy m1 m2 is a by b minus 1 minus x square this is minus 2h by b equal to 0 okay as you can see that x square minus y square 2h by b is nothing but 2xy a by b minus 1 okay first of all 2 terms gets cancelled off okay and here we can write this h by b is equal to xy a minus b by b bb also gets cancelled off that means x square minus y square by a minus b is xy by h and that is a formula which is worth keeping in mind please remember this formula very important formula this is the formula for the equation of the pair of bisectors of a pair of straight lines which is intersecting at origin getting my point okay now next question that will arise in your mind is what if your lines were not intersecting at origin that means they were intersecting at some other point let's say x1 y1 will the equation remain same or will it differ let me ask this as a question to you of course it will differ right because equations do change with shift of origin isn't it dimensions angle between these don't change your shift of origin but of course equation will change with shift of origin correct and if it does how would the equation change let me take this up in the next model so let's say now if your lines were intersecting at origin sorry x1 y1 point and now you want the equation of the pair of bisectors one thing that you would have observed that bisectors are always at 90 degree so if you go back to the previous slide you would have observed that bisectors that is these red lines are at 90 degree correct that means whatever equation I have got for their pair of straight lines equation a plus b for them should be 0 now remember here I am sorry I remember here the equation is actually like this this is acting like your a for this line let me call it as capital A just to not get confused this is acting like your b for this line and this is acting like your 2H for this line as you can see that a plus b would be 0 because 1 by a minus b minus 1 by a minus b has to be 0 therefore our initial theory that we discussed that for a pair of straight lines which are made up of 2 perpendicular lines the coefficient of x square and y square must add up to 0 remember we had learnt a plus b is 0 for perpendicular lines and h square is equal to ab for parallel lines or coincident lines that thing is true over here as well now next question will arise in your mind that if my pair of straight lines are not passing through origin or not intersecting through origin then how would I get the equation of these bisectors for that there is a method which I am going to share with you we first find out x1 y1 point now we will say sir this is a too long a process because then we have to factorize that equation then we have to simultaneously solve those two equations and then we find out the equation of the two lines now see here is an easy method let's say this is given to you let's say this is given to you ok let's say this is given to you you can easily find out the points of intersection by using a simple trick here let me first call this expression by a name s ok guys when we are doing conic section you realize that second degree terms are normally referred to by as s linear terms are referred to by the name as l now x1 y1 could be found out by simultaneously solving dou s by dou x equal to 0 and dou s by dou y equal to 0 solve it simultaneously to find x1 y1 easy this is quite easy I will not give you the proof for this because this is beyond your scope right now but what is the meaning of dou s by dou x dou s by dou x means partial derivative with respect to x when are doing partial derivative with respect to x remember your other variables will be treated as constants of course constants will be constants but other variables other than x would be treated as constant if you are doing partial derivative with y all other variables are than y would be treated as constants ok I will give you a simple example of how this actually works let's say I have given you two lines sorry I have given you this pair of straight lines let me show you how this works I'll come back to this process I'll just show you how it works let's say I ask you find out the point of intersection of these two lines find out the point of intersection of a pair of straight lines given by this okay okay so this is a term which we call as s so first do dou s by dou x equal to 0 what is dou s by dou x this will be 0 this will be 3y this will be minus 6 and everything will be 0 that means y is equal to 2 okay then you do dou s by dou y equal to 0 dou s by dou y will be nothing but 2y plus 3x plus 5 equal to 0 now since y is 2 this is 4 plus 3x plus 5 equal to 0 that means x is minus 3 so these lines will meet at minus 3 comma 2 let me test it out through my GeoGibra so this doesn't mean that dou s by dou x and dou s by dou y gives the two lines no never never never that's what I was about to tell you it doesn't mean that dou s by dou x equal to 0 and dou s by dou y equal to 0 are those two lines by no means okay so I just plot the graph here y is equal to y square okay plus 3xy minus 6x plus 5y minus 14 equal to 0 okay as you can see this is the point of intersection okay let me show you this point this is your point of intersection which point is this you can see minus 3 comma 2 correct this is what we have derived over here minus 3 comma 2 see here and as triple rightly pointed out it doesn't mean that that line is made up of these two lines please both intersection points are common but lines are not the same okay now what next so we have found out x and y1 okay step number one is done step number one is that we have found out x1 y1 next what do you do is in the same equation that we had x square minus y square by a minus b is equal to xy by h there's a reason why I'm writing xy as capital because you're not going to replace your capital X with x minus x1 that is the x1 that you have found out from the step number one y with y minus y1 this would become your new equation for that pair of bisectors basically I have used shifting of origin I have used shifting of origin concept so if your origin shifts in such a way that see basically what was happening your initially point zero zero was here right okay and you wrote this equation right now if you want this point to be called as x1 y1 where should your origin shift your origin should shift to minus x1 minus y1 right that means you have to replace your x with x minus x1 and y with y minus y1 that's what we have learned in our shifting of origin concept that's exactly what we have done so this becomes your answer plain and simple so for this you don't have to find out the two lines you don't have to find out separately the equation of the bisector than all you can easily follow this particular formula and get your answer right so guys we are going to stop over here okay all the best for your chemistry exam which is on Friday okay probably this is our last class before the vacation the shadow vacation begin but don't forget there's a test coming up on 29th of September okay online version 10 to 1 p.m. the same way as you logged in the last time the same login IDs you have to use to the link will be forwarded to you by the rest are in the morning and exactly at 10 to 1 please finish it okay thank you so much bye bye have a good night thank you sir thank you sir thank you all of you