 So last session we started talking about the concept of counting the number of devices of a number. So positive devices of a number and in that we had found out the number of ways or the number of devices of a given number. When I say devices I mean positive devices. I hope everybody remembers the process to do it. It is basically based on your understanding of the combinations from identical objects. So we'll be taking that concept a little further and today I'll be talking about how to find out the sum of positive devices of a given number. Sum of positive devices of a given number n. So for this I'll be starting with a simple example. Let's take a number 24 itself. Okay. Now I want to find out what are the devices and what is the sum of the devices. So you already know the positive devices or the positive factors for this number are 1, 2, 3, 4, 6, 8, 12, 24. Okay. Now, if I want to add these devices, okay, is there any shorter way in which I can do it? In this case, the number of devices are very less. There are hardly eight of them. So adding them to get the answer is very easy. But think in terms of a number which can be quite big and there could be so many factors of that number. So how will you basically accomplish the task of adding the devices? So I'll be taking this number as an example to explain that. If you see 1, 1 is what? 1 is 2 to the power 0, 3 to the power 0. Let me write it like this. This number 1 is what? 2 to the power 0, 3 to the power 0. Correct? Yes or no? Correct? What is this number 3? 3 is 2 to the power 0, 3 to the power 1. Yes or no? Okay. What is this number 2? 2 is 2 to the power 1, 3 to the power 0. What is this number 6? 2 to the power 1, 3 to the power 1. Yes or no? What is this number 4? 2 to the power 2, 3 to the power 0. Correct? What is this number 12? 2 to the power 0, 3 to the power, sorry, 2 to the power 2, 3 to the power 1. Yes or no? What is this number 8? 2 to the power 3, 3 to the power 0. And what is this number 12? 2 to the power 3, 3 to the power, sorry, what is this number 24? 2 to the power 3, 3 to the power 1. Now, when you are adding this, you would realize that from these two terms, you can take 2 to the power 0 common and it will be 3 to the power 0 into 3 to the power 1. Similarly, from these two terms, you can take again 2 to the power 1 common and it will leave you with 3 to the power 0 plus 3 to the power 1. Similarly, from these two terms, if you take 2 to the power 2 common, it will give you 3 to the power 0, 3 to the power 1. And from here also, if you take 2 to the power 3 common, it will give you this. Yes or no? In short, from all these terms that you have, see ultimately, what do we have to do? We have to add all the devices, isn't it? We are finding the sum of the devices. So if you take 3 to the power 0, 3 to the power 1 common from, let me write it slightly far off here, let me write it here. If you take 3 to the power 0, 3 to the power 1 common from each one of these terms, you will end up getting something like this. Correct me if I'm wrong. So this gives you basically nothing but the sum of the devices of the number 24. Some of the devices of the number 24. Let's check whether our answer is correct or not. So if you add them, you will end up getting 1 plus 2, 3, 3 plus 3, 6, 10, 16, 24, 36, 60. Okay, the sum is going to be 60, isn't it? So when you add it, you're going to get the sum as a 60. Okay, let's see what comes out from here. This will give me 1 plus 2, 3, 3 plus 4, 7, 7 plus 8, 15, and 15 into 4, yes, 60. Absolutely. I mean, we are getting the exact result as we desired. Okay, so basically what I'm doing from here, from this small case, I'm trying to see what methodology, what logic I can use to find out the sum of all the devices of any sum of all positive devices of any number n. Okay, so can I extend this concept and make a general formula out of it? Can anybody tell me what is the general formula that will come out from this approach? So let's say we have a number n whose prime factorization is let's say p1 to the power alpha 1, p2 to the power alpha 2, all the way till let's say pk to the power alpha k. Then what do you think is the sum of all the devices of the number, sum of all positive devices of that number? So as per this formula, you take that prime number, you start with the prime number, starting from 0, go till the maximum power that has occurred in the prime factorization for that number. So in this case, I'll be starting with p1 to the power 0, p1 to the power 1, and I will go all the way till p1 to the power alpha 1, just like the way I started with 2 to the power 0, and I went all the way till 2 to the power 3. Same I will do with the other prime factors as well. So p2 to the power 0, p2 to the power 1, all the way till p2 to the power alpha 2, and this process continues till we take care of all the prime factors, that is pk to the power 0, pk to the power of 1, till pk to the power alpha k. Okay, now it may appear to be a very lengthy expression to write, but there is a way to simplify this also. If you see each one of these terms are geometric progressions. Okay, so first term is 1, common ratio is let's say p1 in this case, number of terms is alpha 1 plus 1. So you can write this as p1 to the power alpha 1 plus 1 minus 1 by p1 minus 1. I'm using your geometric series formula. This also I can write it as p2 to the power alpha 2 plus 1 minus 1 by p2 minus 1. And so on till pk to the power of alpha k plus 1 minus 1 by pk minus 1. Please note this down. This is the formula for some of the devices, some of the positive devices of a number. Is it fine? Any questions, any concerns? So everything is coming from that logic that we had used to find the sum of devices of 24. Is it fine? Any questions? See, look at the pattern here, Kartik. Your number 24 was factorizable as 2 to the power 3 into 3 to the power 1, correct? This was the factorization of 24. So what did you do to find the sum till n terms? You started with 2 to the power 0. That is, you took the first prime number, started with 2 to the power 0 and went all the way till the power 3. 2 to the power 0, 2 to the power 1, 2 to the power 2, 2 to the power 3. Same, there is another prime factor, there is another prime number in the factorization which is 3 and it has a power of 1. So same way, start with 3 to the power 0 and go up till 3 to the power 1. What's the point? So let's say hypothetically there was some other number 5 to the power let's say 4. Then what you would have done? 5 to the power 0, 5 to the power 1, 5 to the power 2, 5 to the power 3, 5 to the power 4. I mean hypothetically, I'm just giving a, you know, got it. The same thing I took, let me remove it from here else you'll think that sir has written 2 to 24 as that term which is not correct. Yeah, so in the same way I generalize it, so if the number n is prime factorized as this, I don't know why my n looks like new, okay let me write it again. If I write my n like this, start from p to the power 0, go p1 to the power 0, go all the way till p1 to the power alpha 1, same you do with the other factors as well and just multiply it. Now clear everybody? All right, so let's take a question. The same old question that we had taken, okay the same old question that we had taken, we had I think already done the fact that the number of proper factors of this number, right? Now the second part of the question says find the sum of all these factors, find the sum of all these devices of this number, okay let's do this part only. Of course if you want to give me the answer for both the parts, you are most welcome but I want to know the answer for the sum of the devices. Oh we didn't do this question, okay no worries. So can we do the entire question then, if you haven't done this? So first prime factorize 38808. Please note the question is asking for the sum of the proper factors, the proper devices of this number. Okay, I will also do this question along with you. Now this number 38808, if I'm not mistaken I think most of you would have already prime factorized it. Is it 2 to the power 3 into 3 to the power 2 into 7 to the power 2 into 11 to the power 1? Just check everybody. Everybody's getting this? Okay, great. Now here please understand. The total number of devices, positive devices as I already told you is alpha 1 plus 1, alpha 2 plus 1. Alpha 2 plus 1, alpha 3 plus 1, alpha 4 plus 1, where these are your alpha i's, alpha 1, alpha 2, alpha 3, alpha 4. So take the powers of these prime numbers, add 1 to it and keep multiplying it. So that gives you 4 into 3 into 3 into 2. So what does it give you? I think it gives out to be a 72. But these are total devices. So what are the total number of proper devices? So from proper devices, in proper devices, you don't include the number 1 and the number itself. So you subtract 2 from it. Okay, so there will be 70 proper devices for this number. Is it fine? Any questions? Okay, now I have to find the sum of all these proper devices. So let's first find out the sum of devices. So sum of devices I have already given you a formula. That sum of devices is P1 to the power alpha 1 plus 1 minus 1 by P1 minus 1, P2 to the power of alpha 2 plus 1 minus 1 by P2 minus 1. And this product goes on till Pk to the power of alpha k plus 1 minus 1 by Pk minus 1. Right? So understand here our P1, P2, etc. are 2, 3, 7 and 11. Okay? Or you can use directly the formula also doesn't make any difference. So 2 to the power of now alpha 1 is 3. So 2 to the power of 4. Why am I writing in different colors? Sorry, 2 to the power 4 minus 1 by 2 minus 1. Similarly, 3 to the power of 3 minus 1 by 3 minus 1. Then we have 7, 7 to the power of 3 minus 1 by 7 minus 1. And finally, 11 to the power of 2 minus 1 by 11 minus 1. Correct? These are the terms. Hope I have not missed out anything. Yeah, these are the terms. So let's figure out, this is clear cut 15. This is going to be 13, I believe, because 27 minus 1 is 26, 26 by 2 is 13. This is going to be 48 by 6, 48 by 6 is an 8. Oh, I'm so sorry, not 8. Sorry, this is a 7 cube. Okay. Sorry, what is 7 cube? What is 7 cube? 343, right? Right, Vashna? Yes, what is 7 cube? So 7 cube, 343 minus 1, which is 342, 342 by 6, which is going to be 57. Right? And this is 121 minus 1, which is 120, 120 by 10, which is 12. Okay. How much does this give it to you? I mean, some bit of multiplication is needed over here. So it will give you 133380. Okay. But note, this is the sum of all the devices. I don't need the sum of all devices. I need the sum of the proper devices. Correct? So sum of proper devices will be this minus, this minus, this minus, the number 1 and the number itself, which was 3. Oh, I'm so sorry. Yeah, 38808. 38808. Yeah, how does it come out to be minus minus 1 minus 38808. This comes out to be 94571. Okay, this is going to be your answer. Is it fine? Any questions? Any questions, any concerns? Please do highlight. Okay, we'll take one more questions. Find the sum of even devices, even positive devices of the number, what number should I take? Let's say 1008. Find the sum of all positive devices, positive even devices, sorry, the word even is important here. So even positive devices of the number 1008. Give me a response on the chat box. Nobody? Arav Sinha? Any response? The question of 96. Means the devices sum is lesser than the number itself. F for cafe. Okay, let's see 100 and 1008, how is it prime factorized? What is the prime factorization of this number? 2 to the power 4, 3 to the power 2 into 7 to the power 1, correct? Now, if you want to sum up only even factors, even factor will definitely have at least one 2 in it, isn't it? So in the formula that you learned, in the formula that you learned, you started a prime factor with power of 0, right? But this time for 2, you will start with 2 to the power 1 because at least one 2 should be there, right? Others may start with 0, no problem. But you have to have at least one 2 in a factor to make it even, correct? That's it. And for this, you are taking so much time. I don't know why. Guys, understand the concept. Mugging up is not going to help you out. In the formula, you know that is okay. But you should know what can be the tweaking that can be done to that formula. How can question setter challenge me on that formula itself, right? Most of us, we feel happy solving a problem. No, it should be other way around. You should not be happy solving a problem because once you solve it, you learn nothing from it. So when you solve a problem, make it more complicated. Okay, let's say the question setter had given me the sum of even devices or devices, some of devices which are perfect square. What can I do in those cases? Those kind of thinking process should keep on running in your mind. Okay, anyways, let's calculate this. This is 16 plus 824, 24 plus 428 30. So 30 and this is going to be how much 9 plus 110 into 13 and this is going to be 8. So how much does it come out to be? How much does it come out to be? Yes, so 30 into 13 is 390, 390 into 8 gives you 3120. Okay, so why your answer was 3127 before, Araf? Nikhil, why 3123? Why 3 more than this? Is it fine? Any questions, any concerns here? Any questions, any concerns? See, whenever you are making a factor, your factor is made by multiplying these terms only. So this multiplies with this, multiplies with this and one factor takes birth. Isn't it? That is to itself. Similarly, this multiplies with this, multiplies with this, multiplies with this, multiplies with this, another factor takes birth. So whenever you, what was the reason for using that formula? If you remember, Setu, what did we do? We wrote all the factors in terms of 2 to the power 0, 3 to the power 0, 2 to the power 1, 3 to the power 1, correct? So we made all the factors with the power of that prime number, right? So if I am making any factor, I have to have a 2 in it. So 2 to the power 0, scope is not there anymore. Are you getting my point? Because if you don't keep a 2 in a factor, it will not be even, right? Yes or no? Yes or no? Okay, now think here, how would you find the sum of odd devices? Tell me this. Then I would understand that you have got the concept. What will you keep here, right? So from 2 only 2 to the power 0 can be picked. From others, you can pick any number of terms you want. Okay, so it's just going to be 1 into 13 into 8, 104. Got the point? Okay, so these are the things which are going to be tested while you are asking questions from this topic. Is it fine? Any questions? Any concerns here? Okay, next topic that we are going to talk under devices only is the number of ways to express a number as product of 2 of its factors or 2 of its devices. Okay, now what do I mean by this? Expressing a number as a product of 2 of its devices. Again, let me give an example of 24. Okay, again, let's take the number 24. So 24 is like our frog in the laboratory. So we are doing all the experiments on this 24. Okay, and then we are coming to a generic expression. So we already know the devices of 24. We have already seen 1, 2, 3, 4, 6, 8, 12, 24. So 24 as a number, we can write as the product of its devices in the following way, 1 into 24, 2 into 12, 3 into 8, 4 into 6. Is there any other way? By the way, if you say 6 into 4, it is same as this. Okay, so that's not counted as a different way. Okay, so don't count it again. Don't say 8 into 3. 8 into 3 is same as 3 into 8. Okay, so what is the answer coming out? 4 ways, isn't it? Yes. Now, looking at this, can you make up a generic theory that if I want to express a number as product of 2 of its devices, how many ways can I do it? How many ways can I do it? Okay, what does this example seem to suggest? Okay, it seems to suggest it is half of the total number of factors, right? But yes, Nikhil has a very pilot point. If the number happens to be a perfect square, then this particular approach will fail. So let's say our number 36. Okay, so 36, let's write down the devices first of all. Okay, so 36, the devices, in fact the total number of devices will always be odd here. So 1, 2, 3, sorry, 1, 2, 3, 4, 6, 9, 12, 18, 36, how many? 1, 2, 3, 4, 5, 6, 7, 8, 9. Okay, so there are 9 devices. So can I say my answer is 9 by 2, 4.5? Definitely not. There cannot be a fractional number of ways to do something. Okay. So the formula fails over here. Okay, why does the formula fail is because now let us observe what did we do to get these factors, right? So we started pairing every factor with the other one. Okay, so there was a pairing of one distinct factor with another distinct factor, right? So 1 paired up with 24, 2 paired up with 12, 3 paired up with 8, 4 paired up with 6. Okay. So because there were even number of devices, there was a pairing happening. Okay. But in case of perfect squares, what is the problem? The problem is you can write 36 as 1 into 36, 2 into 18, 3 into 12. See the pairing that is happening. This into this, this into this, this into this, this into this. But 6 has to pair up with itself. Are you getting my point? So there is a pairing of 6, not with some other distinct factor, but with itself, right? And because of that, because of that, your formula doesn't work out. In fact, what works out here is the total number of devices plus 1 by 2. Or what works out here is total number of devices minus 1 by 2 plus 1. Isn't it? Agreed? So basically you're removing that number, which is the square root of that number, then taking all the factors dividing by 2 and then adding that pairing of that square root with itself. So what did I do here? I removed 6. I removed 6 from here. I paired up. That means I did this. And then I added up this guy 6 into 6. So which I'll... Okay. So this one was added over here. 1. Are you getting my point? So either you say number of factors plus 1 by 2, or number of factors minus 1 by 2 plus 1. Both of them will give you the same result. Don't worry. Okay. Is it fine? Any question with respect to what is different that we are doing for perfect square numbers? Okay. So can we generalize it now? Okay. So let's generalize it. So if you have a number n, which is prime factorized as this, okay? The number of phase in which you can express n as product of two of its factors or two of its devices is given by the total number of devices by 2. If n is not a perfect square, n is not a perfect square. Okay. And the same formula, the same question can be solved by using this formula plus 1 whole by 2. If n happens to be a perfect square, if n happens to be a perfect square. Is it fine? Any questions? Any concerns here? Please note this down. And if you have any questions with respect to this, do let me know. So with this concept, we are now moving towards division of distinct objects into groups. We are now going to talk about division of distinct objects into groups. Can you go back? Okay. Okay. Very good. All right. So now we are entering into the concept of division of distinct objects into groups. Okay. So let's begin with the first situation where I'm dividing the objects into groups of unequal sizes. Let me give an example. I have three students. Okay. And I want to make two teams out of it. One team should contain two students and one student in the other team. Okay. So what I'm doing, I'm dividing these students. Of course, students are distinct objects. And you're dividing it into two groups. And you can see the two groups are of equals unequal sizes. One is of size two, another is of size one. Okay. So how many ways can you do it if the order of the group is not important? If the order of the groups is not important. Now, what is the meaning of order of the groups is not important? Order of the groups is not important means if you stop the position of these two, let's say, you know, I, I, I, let's say these are three people, Ram, Shyam, Ganshyam. If I split it like this, Ram, Shyam and Ganshyam, or if I split it up like, you know, I mean, just two groups you're making. Okay. You're just making two groups. Whether, whether one is group one or other is group two, that is of no consequence to us. That is of no importance to us. That means the same two people now are switching their positions. We will not count it as a different division. That is the meaning of order of the group is not important. That means you're just segregating them. You're not giving any name to them that, okay, this will be called group one, this will be called group two. The moment you start calling them group one, group two, then whether Ram, Shyam goes in group one and Ganshyam goes in group two, or Ram, Shyam goes to group two and Ganshyam goes in group one, that would be counted different. But here, my concern is not to assign any order to the group. Just the division is what we are looking for. Are you getting my point? You're not giving any order to this group. That means you're not allowing them to change positions and call themselves as different divisions. Okay, so even if they go like this, I mean, sorry to write it under two, let's say this, I like it separately. It means that even if you do Ram, Shyam and Ganshyam separately, or Ganshyam, Ram, Shyam separately, they will be counted to be the same. That is the meaning of order is not important. Okay, so how many ways can you do this? Can anybody give me a response to this? Okay, so Nikhil, you're just talking about this problem per se, or do you have a logic to solve a bigger problem with respect to the same topic, with respect to the same concept? How did you get that number? That is what I am more interested in. So in order to do this, you just selected two out of three. The moment you select two out of three, the one who is left behind, he will be treated as another group. So a simple way to do this problem is just doing three factorial by one factorial, two factorial. Okay, yes or no? All right, let's say I have another example. Let's say there are five people. I want to make a group of two and I want to make a group of three. Order is not important. How many ways will you do it? How many ways will you do it? Right, so five C2. Yes or no? Correct, Karthik. So it is just as do this. Okay, so can we generalize it? Can we generalize it? So let us say if there are n distinct objects, okay, n distinct objects, and you are breaking up into, let's say, k different groups, okay, where each of these quantities are unequal to each other. That means they are not equal to each other. Okay, so what are you doing? You are creating k groups of unequal size, k groups of unequal size, unequal size. Then how many ways can you do it? So here what you will do is you just keep on selecting the number of items of each group from n. So from n you select first, okay, m items, m1 items, okay, then from the remaining you select m2, then again from the remaining you select m3, and this will continue till you are left with mkcmk. Am I right? This is the generalization of it. Yes or no? And to your surprise and to your happiness, if you simplify this, this will come out to be n factorial by m1 factorial, m2 factorial up till mk factorial. Just like the way it happened in these two smaller examples which I took. So see this 3 when it was divided into two groups of size 1 and 2, it became 3 factorial by 1 factorial, 2 factorial. Similarly here it became 5 factorial by 2 factorial, 3 factorial, and likewise here also it became this. Is it fine? Please note down, this is the result when the order is not important. So we are talking about those cases when the order is not important. That means even if you shuffle these groups among themselves, it will not be counted as a different division. It will still be the same division. Is it fine? Any questions? Okay. So a small question I will take over here, very small question, just as an exemplary question. In how many ways can you divide 52 cards into three unequal size groups of 13, 15 and 24? How many ways can you divide 52 playing cards into three groups of size 13, 15, 24? Please note the question setter will not use the word order is not important. You have to understand. So basically you are just cutting the cards. I hope you understand the meaning of the word cutting the cards. Those who play cards, they use this word. Let's cut the card now. Cut the card means let's split the card. So if you are splitting the cards into three groups, there is no order assigned to it. You can't say that okay, if this card comes here and this card comes here and this card is let's say here, it will be different splitting. No, it will be same splitting. Okay. So how many ways can we do it? What is the answer? What is the answer in this case? Tell me, tell me, tell me fast, fast, fast. 52 factorial, why? Why? Why 51 factorial? Yeah, 51, 52 factorial, why? 13 factorial, 15 factorial, 24 factorial. Is it fine? Any questions? Okay. Now what will happen if the order becomes important? That means let's say, let's say the same question of splitting three students into two groups. If the order is important, then remember, let's say if you do something like this, Ram, Shyam, Ghan, Shyam. Okay. Or Ghan, Shyam, Ram, Shyam. They will be counted as different. Okay. In other words, you are basically assigning some kind of an identity to these groups. Maybe you're calling this as group one, you're calling this as group two. Okay. So if group one has Ram, Shyam and group two has Ghan, Shyam. Or if group one has Ghan, Shyam and group two has Ram, Shyam, then they will be counted as a different division. So the order is important. How would your previous answer get changed? How will this 3C2 get changed? Of course, you'll say 3C2 will get multiplied with 2 or in this case 2 factorial. Isn't it? Yes or no? Same, how will this change? 5C2 get multiplied with 2 factorial. Isn't it? So I just want to generalize it. So in general, if the order is important, the same topic, this is order is not important. But if the order is important, then the formula will change like this. n factorial by m1 factorial m2 factorial till mk factorial, whole multiplied with k factorial. Okay. So this is the result when the order is important. This is the result when the order is important. That means you multiply it with the factorial of total number of groups that were actually there. Why? Because for the same splitting, if I start shuffling these divisions, they will be counted as a different division. As you can see Ram, Shyam and Ghan Shyam, the same splitting, if they were changed like this, then it was a different division when the order is important. Okay. So when normally does it happen, it normally happens when there is a change of hands that is possible of the group or there is a change of the order of the group is possible. For example, for example, let's say there are 52 cards. Okay. You need to split this card as 13, 15 and 24 cards and give it to three players and give it to three players. How many ways can you do it? Now the moment the question says you're giving it to three players, that means the same split can exchange hands. And it can exchange hands in basically you can order these three groups in three factorial ways, isn't it? So because of this part that you are giving it to three players, order becomes important. Order becomes important now. Right? Because this 13, 15, 24, let's say the same 13, same 15, same 24 cards start interchanging themselves. That means let's say earlier this 13 was with Ram, 15 was with Shyam and 24 was with Gansham. Now Ram takes 15, let's say Shyam takes 24 and Gansham get 13. The same 13, the same 15, the same 24. It will be counted as a different way of dividing these elements or dividing these objects. Get in the point. So here the answer will become simple, 52 factorial upon 13 factorial, 15 factorial, 24 factorial, whole multiplied with three factorial. Is it clear? Any questions here? Any questions, any doubts, any concerns? If you have, please do let me know. Great. Now let's talk about equal size groups. Equal size groups. The situation is slightly different with equal size groups. Something wrong with my... Now let's take a situation. Let me start with a very simple situation. Let's say I'm dividing two students into two groups of one one each. And the order of the group is not important. Order is not important. How many ways can you perform this? Okay Nikhil, very good. I want to hear it from everybody. I mean I'm not asked a very difficult question. I have to split two people into two groups. Order is not important. There's only one way to do it, isn't it? Now the answer to this is actually 2C1 by 2. Or you can say 2C1 by 2 factorial. See why it is one and not two Siddharth, sorry Karthik. Let's say Ram and Shyam are these two students. The only one way to split this is Ram is in this group, Shyam is in other group. Is there any other way? Tell me. Is there any other way? So the answer is only one. Alright, let's take this one. If there are four students, I want to divide them into two groups of two students. How many ways can I do it? Tell me. Order is not important. We have not yet assigned any importance to the order. How many ways can you do it? Very good Nikhil. Think carefully and answer. Let's take these people to be A, B, C, D. So in order to split them into two, I can have a team like A, B, C, D. A, B here, C, D here. Or I can have A, C, B, D. Or I can have A, D, B, C. Is there any further split happening or is there any further split possible? I don't think so. There are only three ways. So please note if you're thinking of C, D, A, B, that is only accounted for because order is not important. So here what happened? The answer to this question became 4C2 divided by 2 factorial again and that's how you got 6 by 2, which is 3. Now take a clue from this and tell me that if there were MN objects. So let's generalize it. I mean not a super generalization. There is a super generalization also. So let's say I generalize it to the extent like there were MN objects and you were dividing them into equal groups of size M each. So you are creating N equal size groups. Then how many ways can you do it if the order is not important? If the order is not important, you can do it in MN factorial by M factorial M factorial dot dot dot M factorial written N times. And you have to further divide by N factorial because all these groups have equal size. So when order is not important, this is the formula for splitting MN objects into N groups of equal size. So in short, it becomes MN factorial by M factorial to the power N and further divided by N factorial. So please note this down. So this is the formula when the order is not important. Is it fine? Any questions? If the order becomes important, that means now A, B, C, D and C, D, A, B will be considered to be different divisions. Then what will happen to this result? So if the order is important, let me write it on the top. If the order is important, if the order is important, order means order of the group is important. Let me write it down fully. So what is the meaning of order is important? Order of the groups is important. Then what will happen? Your answer will change to MN factorial by M factorial to the power N by N factorial into N factorial. That means factorial of the total number of groups. So ultimately N factorial and N factorial will get cancelled off. So this is when the order of the group is important. Is it fine? Let's take a small example question on this. 52 cards is divided equally into four groups of 13, 13 each. How many ways can you do it? So how many ways can 52 cards, 52 playing cards, one divided into four groups of equal size? That's part one of the question. Okay. Part two of the question is divided among four players equally. Yeah. Please give me a response to this two questions. Put one and two in your answer while you're giving it so that I understand which part you're answering. Okay. Okay, Nikhil, very good. Very good question. Vashna, excellent, excellent. So the first one when you're dividing it into four groups of equal size, you're just dividing it. You're not assigning it to somebody. You're not allowing them to change hands, right? It is just split. I mean just split them into four pieces. That's it. It doesn't matter whether this split you keep here, this split you keep here, it will be called as a different division. No. So here order is not important. Indirectly here order of groups is not important. You have to understand it reading the question. Questions that I will not mention. Yes or no. And the second method you are dividing it into among four players. So order becomes, order of the group becomes important. So the answer to the first question is 52 factorial by 13 factorial to the power four and further four factorial. And answer to the second part is 52 factorial by 13 factorial to the power four into four factorial into four factorial. Okay. Now, many people ask, why are you wasting time writing both of them? No, there's not a waste of time. You will understand when I give you a super generalization of the concept. But it always doesn't cancel out like the way it is happening here. Okay. So it's a good idea or it's a good practice to write both of them, even though they are getting canceled in this case, write both of them because soon I'm going to give you in fact, immediately after this, I'm going to give you a formula for super generalization of this concept. Okay. Any questions any concerns so far please highlight. Let us say there are an objects. Okay. And distinct objects. Let me write it like this. And distinct objects. You are dividing this distinct objects into several groups. Let's say one group has size M. Okay. Other group has size. Let's say P. Okay. Let me write it like this. P. P. P. Dot dot dot P. So there are K such groups, which have equal size. Okay. And then you're dividing it into let's say R. R. Okay. So let's say there are L such groups, which have got size of R R R. Okay. And let's say I add few more. Let's say one group has size S. Other group has size T. Okay. What I'm doing, I'm making a case where I'm splitting these end distinct objects into some groups. Mind you, few of the groups here have equal sizes of let's say K of these groups have equal sizes of P. L of these groups have equal sizes of R. Okay. So this is a super generalized version. So how many ways can you make this division if the order of the groups. If the order of the groups is not important. Okay. Now listen to this. If the order of the group is not important, you'll take total number of items divided by factorial of this. And here T factorial S factorial M factorial. And these P factorial, you will write K times. So it's better to raise it to the power of K. And since there are K groups of equal size, you have to further divide by K factorial. Okay. Similarly, here are factorial. You will write L times. And since there are L equal size groups, you'll divide further by factorial. Are you getting my point? What I'm trying to say. For example, let's say I'm dividing 52 cards as 555. Let's say six, six, by the way, how many over 15 plus 12, 27 over now. So 27 over. So I have 25 more to go. Right. So one is 10. Let's say other is 15. Okay. So you just have to divide these 52 cards like this. Three groups of equal size five, two groups of equal size six and two separate groups, one of size 10, another of size 15. If order is not important, that means you're just dividing them into group, not giving it to any player or something like that. Then the answer for this will become 52 factorial by, tell me, five factorial cube because 555 three groups are there and further divided by three factorial because they are three groups of equal size. Are you getting my point? This is very important. Don't forget this three factorial. Okay. Similarly, six factorial to the power two, further divided by two factorial because two equal size groups. And then 10 factorial, 15 factorial. Is the idea clear everybody? Why? Because they were three equal size groups. Why did you divide by two factorial in the beginning? Or when you were doing, when we're splitting it up into equal size groups by the same logic. So this term will come because there are K groups of equal size. This term will come because there are L groups of equal size. Got the point. Similarly, here also you'll divide by three factorial because three groups are of equal size. And again, two factorial here will come because these two groups are of equal size. Got the point. Okay. Now let's talk about when the order of the group is important. How will the same problem will be solved? Order of the groups is important. Okay. In this case, your previous result will just get multiplied with the total number of groups factorial. That's it. Okay. Very same formula. N factorial. Let me write it in yellow. N factorial by, what did I write? I'll just copy it. T factorial, S factorial, M factorial, P factorial to the power K, K factorial, R factorial to the power L, L factorial. This will get multiplied with the total number of group factorial. How many groups are there? One, two, three, K, L. So K plus L plus three factorial, isn't it? So K plus L plus three factorial. In short, I mean, I've just wanted to write here. It's the total number of groups so that whatever is the situation you can adopt, you can adapt to that situation and write it down. Yes. And in this case, it will not get cancelled. And in this case, it is not going to get cancelled. So let's take the same example. How many ways can you divide 52 cards? 52 cards among seven players says that three of them get 555. Two of them get 66 and other to get 10 and 15 respectively. So how would the answer change here? So now you're dividing it among players. That means order is important now. So if order is important, it will become 52 factorial upon five factorial cube by three factorial, six factorial square by two factorial, 10 factorial, 15 factorial and into seven factorial. And this is where the cancellation is not going to happen. Unlike what happened in the previous case. Clear everybody. This is very important. Unfortunately, in CBC curriculum, they don't speak about this concept. It is very well stated in ISC books, ISC curriculum. Okay. Should we take up questions now? Let's take questions. I think we have already done the first two. You may probably start with the third part here because I've already done. How many ways can a pack of 52 cards be distributed equally among four players in order? And second part divided into four groups of 13 cards. These two we have already done. In fact, I will write down the answer also quickly so that we don't waste time. Okay. So this is the answer to the first question. Answer to the second question was this. I want you to solve the third part. Dividing into four sets, three of them having 17 cards each and the fourth just having one card. Okay. Now please read the question carefully. They're not saying dividing equally into dividing it between the among the players or giving it some kind of an order. So they're just dividing them into groups. So order is not important here. That means even if you change the position of these divisions, it is not going to be counted as a different division. Excellent, Nikhil. So you're dividing 52 like this, 17, 17, 17 and one, isn't it? So it is going to be three groups of 17. So 17 factorial to the power three and also three factorial will come in because three groups are of equal size. Okay. So can I write one factorial also? Write it no problem. It doesn't make any difference. One factorial is anyways. Got it. Is this the answer you got, Vaishnav? Yes, then you're right. Okay. Can you take another one now? Very good, Karthik. Can I take another one now? Any questions, any concerns? All right. So let's take another one. So the number of groups having equal number of objects, you have to also divide by the factorial. Why didn't you ask that? Why when we were generalizing it? Why did you divide this by two factorial here? Let me go back. Let's start in the beginning. I think it was here before this. Yes. So why did you divide this by two factorial? Why did you divide this by two factorial and generalization here? Why did you divide this by n factorial? What is the answer for that? I mean, I understand you're asking this question very late, but you should have asked it, you know, then. See here, what is happening? When you are, let's say I take this case, this is a more, you know, interesting case. Yeah, this one. I'm just putting it into two parts. Please note that the size of the group is equal. That means there cannot be any distinction made with respect to the groups with respect to the size, different size themselves attribute to different groups. Correct. Anybody can understand that they are two different groups if their size is different, isn't it? Yes or no. But in this case, the fact that the groups have equal sizes. That distinction that the group are different with in view of their size is taken away. See, many times what happens, it is very easy to understand that these two are different size groups because the number of items there are different in each one of them. They are unequal size. So they're different groups. So the size of the group acts like a distinguishing. Isn't it? So here, that factor is taken away from this because the size of the group doesn't make any kind of a distinction. In earlier, there was Ram, Shyam, Ghan, Shyam. Correct. So if I put the Shyam on the other group, it was, it is going to be counted as a different thing. Right. But here what is happening? Let's say in the first group again, CD comes and AB comes in the other group. It will not be counted as a different splitting. So the virtue or the advantage of size difference is now taken away from these. So your answer gets reduced because the sizes themselves were making a difference in their and the number of the counts. So if the size is different, then automatically the counts will increase. But if size is equal, what will happen? Size advantage is gone. There is no size advantage given to these groups. And that's the reason why your answer drops by a factor of the factorial of the total number of groups having equal size. Are you getting my point? So size difference was one of the factors which led to more count. But in equal size, that advantage is lost. Size difference is no benefit. Are you getting my point? So they're all equal size. So what will happen is, even when you are splitting, let's say you took A and B out. Okay. And you allowed the other two, C and D to stay there. Next time you take C and D out and allowed A, B to stay back. Now because they were of the same sizes, A, B, C, D and C, D, A, B will not be counted as a different entity or not be counted as a different division. Do you understand? Right. So, for example, you chose two out of four people. If you choose A, B, then C, D is saved. Okay. If you choose C, D, then A, B is saved. But are these two different counts? No. Right. They are not different counts because A, B, C, D, C, D, A, B will be same because of the group size. Correct. But let's say there were three people or let's say there were four people and you took one out and the three remain back. This will be counted as a separate because if you take one and three remain back, they automatically became two unequal size groups. So there cannot be a repetition of this counting again. But in A, B, C, D case, there would be a repetition happening. That is the main reason why we are dividing it by factor. But you should have asked these questions in the beginning. Right. Many people don't ask this question in the beginning and then they later on realize that why are we dividing it by factory? Okay. Coming back to this question. In how many ways can 12 different balls be divided between two boys one receiving five other receiving seven. And in how many ways can these 12 balls be divided into five, four and three balls respectively. So there are two questions into this. I'll call them as one and two. So this is your second part. Let's say this is your first part. Let me write it in a color which you can see. This is the first part. This is the second part. Yeah. So first part you are giving this 12 balls to two boys. Okay. So you can think as if there are two players between which you are splitting it. So order becomes important here. Okay. So what is going to be your answer very good area of very good tradition. Very good business vara first one. So sol factorial by five factorial seven factorial into two factors. Are you getting your point? If you're not allowing it to change hands. Then you don't multiply the two factor but here you don't know, who is going to take how many words. can get five, Shyam can get seven or Ram can get seven, Shyam can get five. Got it? Okay, next. In how many ways can you divide 12 balls into five, four and three? Just division. So order here is not important. Okay, so your division is going to be 12 factorial by five factorial, four factorial, three factorial. Is it fine? Any questions? Siddharth? No. Vashnav? No. Second part, please see carefully what mistake you're doing. Vashnav, why are you further dividing it by three factorial, sorry, why are you further multiplying it into three factorial? Do you know whom these, do you know that these have to be given to some people or order is important? Then why are you multiplying with three factorial? Same goes with Siddharth also. So don't multiply with three factorial, order is not important. Had I said that I'm splitting 12 balls as five, four, three and giving it to three students or three players, then I would multiply with three factorial because the same five, four, three can switch hands now. So that's switching of hand if it is allowed and you are counting that as a different way of distribution, then multiply with three factorial, else not. Okay, so let's take this one now. Okay, Vashnav, the kill, very good. Okay, 16 different books are to be distributed among three players such that A gets, sorry, B gets one more than A, C gets two more than B. Okay, good. Okay, Siddharth. Okay, say just funny, very good. Okay, so let's say A gets X objects. Okay, so B should get how much, X plus one, C should get how much, X plus one plus two. Okay, so altogether it should be 16, that means three X plus four should be 16. That means three X is 12, so X is four. Okay, that means A should get exactly four, B should get exactly five and this C should get exactly, exactly seven objects. Okay, so the answer here would be 16 factorial by four factorial, five factorial, seven factorial. Sir, are you missing three factorial? No, if you're writing three factorial, it will be wrong. So Karthik Sanoj, Setu, see think, why are we multiplying with three factorial? Does this, does this four, five and seven allowed to change hands? Tell me, can A get seven objects, B get four objects and C get five objects? Tell me, can it happen? Are these objects allowed to change hands? Does order, will order be taken into account here? Then why are you multiplying with three factorial? That's the whole thing I'm speaking from, you know, since the beginning of our session, don't mug up things without understanding it. We multiply it with three factorial. If the same four object, five objects, seven objects, can be shuffled among themselves. So Ram gets four, Shyam gets five, Dhan Shyam gets seven. Let's say the same four object that Ram got, now it is given to Shyam. The same five object that Shyam got was given to Ram and let's say Dhan Shyam retains his object. So in order to account that these will be counted as a different way, we multiply with three factorial because there are six ways in which they can change the, the same objects, the same division can exchange hands, isn't it? But here there is no scope of exchanging hand. Ram has to have only four, he can't get five, he can't get six. So it is as good as order is not important despite there are people involved in this question. So apply your brain when you're solving it. Don't just think, oh, people are there, Ram, ABC. Let's multiply with three factorial. Are you getting my point? Okay. So I think R of Sinha got it right and Tejaswini got it right. These are the only two people who got it right. Thoughts are nothing, it should be on papers. Okay, good. I think we had a good exercise on division into groups. Awesome. Now we're going to another sub topic, arrangement of and distinct objects into our different groups or our distinct groups, whatever you want to call it, different distinct. Okay. Now, see here, the word in this, you know, name of the topic that makes lot of change is arrangement. It is not a distribution, it is not division, it is arrangement. What is it? Arrangement. So how many ways can you arrange and distinct objects into our different groups is what we are going to talk about. Can somebody tell me what is the difference between distribution and arrangement or division and arrangement? What is the difference between division of and distinct objects into our groups and arrangement of and distinct objects into our groups? Order of what? Groups matter. That is already taken into care. Our different groups means order is important. Order of group is important. Let me write it in white. Right, Setu, absolutely right. Now here, not only order of group is important. Okay. But there could be ordering of objects within the group also possible. Okay. Up till now, we did not care whether the split was Ram, Shyam, Ghan, Shyam or Shyam, Ram, Ghan, Shyam. That means within group 1, if you change Ram, Shyam to Shyam, Ram, it was not considered to be a different division so far. But now in this topic, we are going to take that also into our consideration. Getting my point? Let me give you an example. Let's say there are two boxes. I mean, I'm calling them as group 1, group 2. Okay. And there are three balls, distinct balls, ball 0 1, 0 2, 0 3. Okay. So if I say, arrange these three objects into these two groups. So let's say you put two here, one here. Or if you put the same two like this, they will be counted as different arrangements. Are you getting my point? But they would be counted as same distribution, but they will be counted as same division or distribution, division slash distribution. There is a difference between arrangement and division slash distribution clear to everybody. Is there anybody who still has a doubt? That means within a group, if you change the position of the objects, it will become a different arrangement. Okay. That is what we are going to talk about in this topic. Okay. All right. Now let me give this as a food of food for thought for you people. How many ways can you arrange and distinct objects into our different groups? If empty groups or blank groups are allowed, blank groups or empty groups, what do we call it? Empty means it. Any group can be empty also. Groups are allowed. How will you solve this question? Think and answer. So think as if there are n different balls and you have to place them into our different boxes. How many ways can you do that? Not only place, you have to arrange in our different boxes where you can also have a group which is blank. How will you do it? Theory won't be discussed for everything like this. I mean, we are not trying to mug up all the types of concept. Think as if this is a question and you have to solve it by your existing knowledge, by your prior knowledge. How will you do it? Don't always expect spoon feeding. Okay. NPR, N minus 1 PR minus 1. Sir. Yes, Abhijej. Tell me. Sir, I have some doubt. So sir, like currently my terminal exam is going on and this Sunday is a game on test. So can I take the test on some other day also? Yes, yes. Why not? If you have an exam coming up, you can always take it the other day. So sir, do we have to inform? No, it's okay. We come to know who are taking the test. We also come to know who have not taken the test. You can take it on some other day also, not any. Okay sir. RPN plus 1. If Karthik, R is less than N, then your answer goes for a cross. Okay, Nikhil, I would like to hear from you. Why do you think it's N factorial into N plus RCR? Any special reason for that? I mean, you can unmute yourself and talk. Yes sir, because you first arrange the N objects in N factorial ways. Okay. And then basically you make it N plus R places and you put it. If you want to divide into R groups, you need R minus 1. Yeah, N plus R minus 1. Okay. Yeah, then C R minus 1. Yeah, my formula is wrong. No, in this case, there will be, see, why did you do N plus R minus 1, C R minus 1? Because you need like on the left side of the first dash you put, those many stuff are going to group 1. Then between the first dash and the second dash, those many stuff are going to group 2. And second dash and third dash, those many stuff are going to group 3. So you'll have N minus 1 dash in the end. Like with at the end group will go on the end group. Okay, I mean, I understood your concern. Can I solve your question in this way? See, as, as Nikhil rightly said, in order to divide N objects, okay, into R different groups, you would require R minus 1 sticks or partitions. Correct. Let's say I take these R minus 1 sticks. Okay. And remember sticks, when you say R minus 1 sticks, they will become identical sticks. How many people question me here, sir, aren't you dividing it into R different groups? So why are we taking identical sticks? So sticks are, sticks being identical doesn't stop the groups from being different. Okay. It just says that if you place a stick, there is a division made by the stick to the right and the left of the stick. Right. For example, let us say I want to divide five objects into three groups. Okay. I'm taking a simple example here. So O 1, O 2, O 3, O 4, O 5 are five distinct objects. And I want to divide it into three groups. Right. So what I do, I take two identical sticks and just place them here. Let's say I place them here. Okay. So automatically this becomes group number one. Automatically this becomes group number two and automatically this becomes group number three. Are you getting my point? So identical sticks do not stop the group from being distinct. Okay. Please remove this false notion in, in, in your mind if at all you have any. Okay. Now see coming back to Nikhil's thought process. So Nikhil is trying to basically say that first of all, we arrange these n factorial objects among themselves. Okay. And then what he is doing, he is trying to make an imaginary place. Okay. Imaginary seats of n plus r minus one. Let's say n minus r minus one chairs. Right Nikhil? Yes sir. Okay. And now he is doing, he is, he is basically making the, making the, I mean, I hope you are making the stick sit first. Right. Okay. Yes sir. I'm making the, making the sticks sit first. Yeah. So he is basically choosing these many sticks and making them sit. Okay. And remaining places, the balls will already come and they already arrange themselves among each other. So do you think this is going to give the right answer? In my opinion, this is absolutely right. Okay. In other words, the answer that Nikhil you will be getting will be n plus r minus one factorial by n factorial r minus one factorial. Okay. n factorial, n factorial will get cancelled off. Okay. Now this rings another bell in my mind when I see this result, of course I've used Nikhil's result only, but I'm going to think this from a different perspective now as what Nikhil had done. Now what I'm going to do, I'm going to mix these objects and these sticks. So these n objects and these r minus one identical sticks, I'm going to mix. Okay. And I'm going to arrange them in all possible way I can. Right. How many ways can you arrange n plus r minus one objects of which r minus one are identical? What is the answer for this? If you recall your Mississippi question that we did in the other class, it is total number of objects factorial by factorial of those many objects which are identical. Do you see that my answer has come out to be same as what Nikhil has got? Yes or no. Right. So it's all about thinking process and end result will come out to be correct if your logic is correct. Okay. But very well done Nikhil. That's a good thought process. Is it fine? Any questions, any concerns with either my way or Nikhil's way? Do I like, please? I'm getting some questions. Could you explain again whose way, my way or Nikhil's way? My way. Okay. See what I did was, okay, let me take a simple example, Karthik, then you will understand. Let's say I have five objects and I want to distribute them into three groups. Okay. Sorry. I want to arrange them into three groups. I keep forgetting that I'm doing arrangement not distribution. I want to arrange them into three groups where blank groups are allowed. See here, the question is blank groups are allowed. So you can have a group which is having no item at all. So what I did, I took these objects and I took two sticks. Okay. So altogether I have now seven objects of which two are identical. So two are identical in this. Which two? The two sticks. And I started arranging them. Okay. And I started arranging them left, right and center. Okay. So whatever arrangement I make, that will lead to an arrangement of these five objects into three groups. How? Let's say I end up getting it, you know, having an arrangement like this. Stick O1, O4, stick, O3, O2. Anything is left? O5. Okay. Now see, it automatically means group number one is blank. It automatically means group number two has these two items O104. Group number three has got O3, O2, O5. So since you're allowing the objects also to be a part of the arrangement, there would be arrangement of the objects within the groups also in this way. That means, let's say if I had a case like this where there was a stick, there was O4, O1, stick, O2, O3, O5. Then as you can see, for the same split that is O104 in the second group and O2, O3, O5 in the third group, since their arrangement of these intra group objects are changing, they are basically counted as a different arrangement for us. And this is accounted for in the formula seven factorial divided by two factorial. Do you remember the Mississippi question that we did? What did we do? Total number of objects divided by factorial of that many objects that we did? Okay. Yes or no? Okay. So there's just many. Can we solve it without using the idea of identical sticks? Okay. What else can you use? You suggest? No. Now, Nikhil's method was what Nikhil did this thing. Nikhil made seven places. Okay. Because he knew that total number of objects plus the sticks will make seven, requires seven places. Right. So he's now placing these distinct objects and these sticks on these places. Okay. So what he did first, he chose two places where he wants to place his sticks. So let's say he chose, he chose this place and he chose this place. Right. And then what he did, he took these objects, the remaining five objects and distributed it in the remaining places. And not only he distributed it or not only he put it, but he arranged them also. So that can be done in five factorial ways. Yes or no? So when he did that, so let's say O1, O4 comes here, O3 comes here, O2 comes here, O5 comes here. He automatically got seven factorial by two factorial, five factorial into five factorial. And as a result, it became the same as what I discussed. Clear, Sethu? Okay. Now, the second part of my discussion here is how many ways can you arrange and distinct objects into our different groups if blank groups are not allowed, blank or empty, whatever you call it, doesn't make a difference, blank or empty groups are not allowed. Let me write not in caps, not allowed. Yeah. A bold statement, not allowed. Yes. How many ways can you do it? Again, think. Just like Nikhil thought for the previous one, here also you think. You should be getting the answer. And I will ask you to defend your answer. Okay. Blank groups are not allowed. Very good. Excellent, Nikhil. Okay. Prishan, good. See, now in this case, the situation is slightly different from the previous one in the sense that now these r minus one sticks that you have, these r minus one identical sticks that you have, they can only be placed in the gaps between the objects. That means wherever I'm showing you the crosses, only there you can place these sticks. These sticks cannot come in front. Right? Not even one. So none of the sticks or sticks should be able to precede the first or succeed the last. Or not only that, they cannot come consecutively in between also. Why? If they do so, blank groups will get created. No. Are you getting my point? So let's say if I place the sticks like this, stick, stick, o one, o two, o three, stick, stick, stick, o four, o five, I mean, whatever, then what will happen? This group will become blank. Sticks are acting like partitions between the groups. So this is first group blank, second group blank, not allowed. Here also blank, blank, not allowed. So you can't allow your sticks to come either one or more in number in front of the first object and in the last of the last object or after the last object and consecutively in between. So you can only place your sticks strategically in between these gaps only. Okay. So first, let me first arrange the or put these r minus one sticks in these n minus one gaps. All of you please confirm this. How many gaps will be there? There will be n minus one gaps between the objects. Okay. Choose any r minus one and place your sticks over there. So how many ways choose r minus one gaps from n minus one and place your sticks there. So that will be n minus one CR minus one. But you can also arrange the objects also in the very same placement of the sticks. That means objects arrangement can also be done. Okay. So n objects can be arranged in n factorial. So altogether you have to take both the events into your picture. So your answer will be n factorial times n minus one CR minus one. Is this fine? Any questions? Any concerns? Absolutely right, Nikhil. Your answer is perfect. Anybody who has any issue in understanding this logic, please let me know. Please highlight. Formula. Formula for the previous one was n plus r minus one factorial by r minus one factorial. Fine. Okay. Let's take a small question on the same. How many ways can five different balls, how many ways can five different balls be arranged into three different boxes so that no box remains empty? Yeah. Sure, Karthik. I'll go back to the previous slide after this question. Everybody, please give me a response in the next 20 seconds. Not more than that. You already know the formula. Hey, you missed out something. Into some factorial you are missing, Nikhil. How many ways can five different balls be arranged in three different boxes? No box remains empty. Correct. No, five factorial only. Yes, only Nikhil has given a response so far. What about others? Quick, quick, quick, quick. No box remains empty means blank group is not allowed. Very good, Tejasini. Correct. Anybody else? Very good, Karthik. Yeah. So the answer that will be obtained for this is n into n minus one CR minus one. n is obviously five, r is obviously three. So use this formula. But again, don't rely on formula. It's better you apply your brain first. Okay. So it's 120 into six, 720 ways. Is it fine? Any questions? Anybody? Any questions? Okay. Let's say I remove this factor. Now, let's say box can be empty. Empty boxes are allowed. Then how will the answer change? Empty boxes are allowed. I mean, we sometimes also call it as arrangement without restriction. That means let there be blank boxes also. Let all the balls come in one box also. No issues, no restriction at all. So how would the answer change? Okay, Manu. So again, think as if five balls and two sticks because there are three boxes and two sticks. Together, how many arrangements you can make? That is your answer. Just remember this. So what is the answer? Seven factorial by two factorial, right? Isn't it? Seven factorial by two factorial. Any questions here? So answer will be 5040 by two. It's almost 2520, 2520 ways. Is this fine? How can the answer be lesser than when there is a restriction? Without restriction should be more number, isn't it? So it can't be 5C3. No, Manu. Okay. Is it fine? Any questions? All right. Now, the next concept that we are going to talk about is very similar to what we have already done before this concept. Division into groups. Oh, yes, yes, yes. Sorry, sorry. So, Kartik, one second. Yeah. The one before this, where this side, is it the right position of the slide or you want me to drag the slide somewhere else? Done. Okay. Thanks. Which ball question? This one only. See, let's say you have five balls and you want to divide this into three groups using two sticks and six can only come in these positions. So choose any two positions where you want to put the sticks. That's it. They cannot be shuffling of the sticks among themselves because they're identical. But there could be shuffling of the objects among themselves that is five factorial. Why do you choose R minus one stick? That's a very, very basic doubt you have asked. Okay. If you want to cut a thread into two pieces, how many cuts do you make? Have you ever cut a thread into two pieces? How many cuts do you make to break it into two pieces? One cut, no. So one partition is sufficient to create two groups. Correct. How many cuts will you make to make them into three pieces? Two cuts or two partitions will be required to break them into three pieces. So if you want to create R groups, you need R minus one partitions or R minus one sticks. The next concept that we are going to talk about is distribution or division. Distribution or division of end distinct objects into R different groups. Now these concepts sound very similar to each other. First of all, distribution and arrangement, people think they're same. They're not actually. Now, when somebody, when you read this, you'll say, are you, sir? We did this. Didn't we do the division of objects into groups? Isn't it the same thing? Of course, there are many things which will be similar, but there will be some certain things which will be new to this concept or very pertinent to this concept. For example, when we were dividing the objects into groups, was there any case where some group had zero items? No, right? Now this concept, we will have zero items. Okay. In the previous discussion, our groups order may not be important, but here order is definitely going to be important. So this, this you can think to be more robust. Okay. As compared to the previous one. So here even the blank groups can be allowed. That is the difference between this concept and the previous concept. Okay. So let us try to discuss this before our break. So I'll take, I'll divide this topic into two cases. The first is where blank groups are allowed. One groups are allowed. Now, mind you here, you're only doing distribution. Okay. No arrangement. So how many ways can you distribute and distinct objects into our different groups such that blank groups are allowed? To answer this question, think like as if there are R boxes, B1, B2, B3. These are our different boxes and there are N different walls. Okay. There are N different objects. How many ways can you place these N different objects into these boxes such that there is no restriction at all. That means blank group is allowed. We normally call this as no restriction case. This is actually a question which you can solve from your understanding of your first class only. Very simple. Absolutely right, Nikhil. See, answer comes like this, fraction of a second. Conceptually, you should be sound. That's it. Come on. I'm waiting for others also to respond. Just one person responding is not, I don't feel happy with that. I feel happy when multiple people get right. Yes, they just need to try. Manu, Karthik, Karthik Sanoj, Setu, Prashim, Pashnav, Arav, Sharvili. Okay. This is a super easy question, but I think other than Nikhil, nobody is getting this. See, ask yourself, every ball will have how many options? How many options this guy has? Our option, this guy can place himself in any one of the R boxes. How many options this guy has? Many people here think one box can accommodate only one ball. No. Any number of balls can go to any box. Right, Tejasini has now given the answer, but let's complete this. How many options O2 has? That also will have R. This also will have R. This also will have R. This also will have R. So as per fundamental principle of counting, the total number of ways will be R into R into R. That is R to the power n. So the number of ways in which you can distribute n balls and distinct balls, let me correct myself, distinct balls into R different groups is R to the power n. This will also include cases where some boxes may also remain empty. Okay. That means no restriction at all. Any number of balls can go in any number of boxes. Sorry. Any number of balls can go in any number of box or any box and any number of boxes can remain empty also. Is it fine? Good. This was the simplest one, but the other one is slightly tricky, which I'm going to take it in the next page. How many ways can you distribute these n distinct balls in these R different groups if blank group or empty group is not allowed? Is not allowed? Okay. Think and answer. Think and answer. Could you go to the previous slide? Yes. Why not? Prisham, any question? Any discomfort with respect to the result? Do let me. Or for that matter, anybody? Anybody who wants to challenge the results, please, you're most welcome to do it. So if you feel no, sir, I feel there's a different way to do it. This is not going to work or this is going to give me erroneous results. Please bring up such kind of discussions and debates that will add more value to the class, rather than being a monologue from my side. Okay. I'll come back to you. I'll ask the reason for that. Let me hear it out from others as well. Okay, Nikhil. The answer that is going to come out for this is going to actually blow your mind off. It's slightly complicated, but I will first tell you the normal approach which people take to solve this question. Okay. Then I'll tell you what is wrong in that normal approach. See, I have to distribute N different balls into these R different groups. Okay, there are N different balls and there are R boxes. B1, B2, B3, dada, dada, dada, till BR. Now I don't want any group to be empty. So what do I do? I pick one, one, one ball and put it in these boxes first of all. Correct? So let's say I choose R balls from these N balls. Correct? And I place them in these boxes. Correct? Yes or no? That means I can place them in R factorial way. That means every ball has, sorry, every box has got one ball. Correct? Now what do I do? The remaining N minus R balls, each one will have RR options each. So can I say it will be R to the power of N minus R then? But this result is wrong. Can somebody tell me why it is wrong? Okay, Nikhil has raised an immediate objection. Nikhil has said you will be double counting things. Okay, Nikhil, how I will be double counting? Can you explain? At least can you explain to everybody if you want? You can unmute yourself if you want to. So when you first take the R objects and put in the, when you first take the R objects and you put in the R groups, then you take the rest N minus R objects and then put it in. So that will also be taking one of the N minus R objects and then putting it in and then taking, so you will be double counting something. So I do not explain it. Okay, I'll help you out. See what Nikhil means to say, I'll give a simple example. Let's say out of these R objects which I selected, I selected, let's say ball number 2. Okay, and I put ball number 2 in box 1. Okay, and let's say in these N minus R objects, there was ball number 10, let's say. I'm just assuming. Okay, and then this 10 ball, while I was distributing, when I was doing this activity, this ball number 10, I put it in again box number 1 only. Okay, so let's say box 1 has finally O2 and O10. Now let's say in this NCR, I happen to choose ball number 10 now. And I put it in this box B1. And in this N minus R, I chose ball number 2 now and I put this again in box number B1. And what I did, I counted them as a different way of doing it, but in reality, are they different? No. O2 and O10 coming in box 1, let's say, whether O2 comes first and then O10 comes later on or O10 comes first and O2 comes later on is the same distribution. So there will be a double, in fact, there will be a multiple counting which will happen in this result. That is why this result is wrong. Are you getting my point? Nikhil, am I in sync with whatever you were thinking? Yes, sir. Okay, so this is an approach which I have seen many students taking. This is why can't we do this problem like this? So in this approach, there is this problem of multiple times counting. Your answer will come much more than what it is supposed to be. Are you getting my point? Okay, so this failed. So how else we should do this question? So for that, I will be using a very well-renowned principle called principle of inclusion exclusion, which you are already aware of when we were doing this chapter of sets, sets relation functions. And then the sets part, we did the cardinal number properties. And in one of the properties, we had talked about principle of inclusion exclusion. If you remember, NA union B is NA plus NB minus NA intersection B itself is a principle of inclusion exclusion formula. You can extend it to any number of sets you want. So include one at a time, exclude two at a time, again include three at a time, exclude four at a time. So inclusion exclusion, inclusion exclusion keeps happening. So that principle I am going to call up in this particular concept and try to address the situation. Let's see how all of you please pay attention. This is a slightly new thing for you all. So I'll erase this off. Okay. So let us say, first let me write, I will use principle of inclusion and exclusion. Inclusion and exclusion. P IE normally we call it principle of inclusion and exclusion. Okay. All right. So here I will define certain events for you. Listen to that events very, very important because using those events or using those sets, I will be solving this question. Why the previous formula was wrong? Okay. Situ will do one thing. We'll connect offline. Okay. Because we are towards the break side also and I have to finish this off. You call me and I will explain it to you. Okay. Same example. Or maybe after the class you can stay back. I'll explain it to you again. Now all of you please pay attention. I'm going to solve this question by defining certain events. Let me just remove this because so let us say I define an event E1 where even is an event where box B1 is empty. What I've done. I'm defining an event. I'm calling that event as E1 event or set. I'm defining a set where this set contains all those cases where B1 is going to be empty. Okay. Now tell me in how many ways can I distribute the balls such that B1 box remains empty. Others may or may not be empty. Don't get me wrong. Many people think said only B1 is empty. No. B1 is empty and only B1 is empty. There are two different words, two different sentences. I've not said only B1 is empty. I've said B1 is empty. Others may or may not be empty that I'm not trying to say here. I'm just saying B1 is empty. How many ways can I distribute n distinct objects into these R boxes such that no object goes in B1. Very good Nikhil. Anybody else? I think as if every ball has got R minus one choice now. O1 has R minus one choice because B1 can't go in. It cannot go to B1. O2 also has R minus one choice. It cannot go to B2. So when you're saying R minus 1 to the power n, basically other boxes may or may not be empty. But B1 will definitely be empty. Got the point? Any questions? Okay. Similarly, let's say I define an event E2 and not be writing too many things where B2 is empty. So tell me what is n E2 then you'll say, sir, come on. Same thing. R minus one to the power n. Now this time, you're not allowing any ball to go to B2 box. But still R minus one boxes are there and it can choose any one of the R minus one boxes. So can I see if I take EI to be the event where box BI is empty, then the number of ways in which I can perform this event is same R minus one to the power n. Everybody's happy with this explanation? Okay. Now a similar set of event I will make even intersection E2. So let's say even intersection E2 is an event where both box B1 and B2 are empty. Now I do not mean to say others are full. Others may also be empty. Okay. May or may not be empty. Others may or may not be empty. But B1, B2 is definitely empty. So how many ways can I perform this activity where B1 and B2 is definitely empty? Right? So R minus two to the power n. So I'll just quickly generalize it. So if I define an event like EI intersection EJ where this event says BI and BJ are empty, then the number of ways to do this activity will be R minus two to the power n. Okay. In general, I'm not taking a lot of time. So similarly, if I say let's say EI intersection EJ intersection EK is the event where these three boxes BI, BJ and BK are empty. Okay. Rest may or may not be empty. Then the number of ways in which I can perform this activity is R minus three to the power n and so on. Duck, duck, duck, duck and so on. So you can have now four, four boxes empty, five boxes empty, six boxes empty, et cetera, et cetera. And accordingly, your answer will change. Okay. Now some of you will be wondering, are you, what does sir want to do here? I'm not able to understand. Now let me connect the dots. Let me give you now the complete picture. What are we looking for? We are looking for those cases or the number of those cases where box B1 is not empty and box B2 is not empty and box B3 is not empty till box R is not empty, isn't it? This is what we need. This is the need of the question, isn't it? What does the question say? I have to distribute the balls in such a way that empty boxes are not allowed. So I have to count those events where box B1 is not empty and B2 is not empty and B3 is not empty and BR is not empty. Yes or no. Yes or no. Now as per de Morgan's law, this expression, can I not write it like this? Which law? De Morgan's law. I'm sure you remember, we have done this in our sets chapter, de Morgan's law. And not only that, by using your cardinal property, you can say N universal set minus N, E1 union, E2 union, till ER. Again, from your cardinal number properties, we can say summation 1 at a time, that is N, E1, N, E2, till N, ER. Minus summation of intersection 2 at a time. See, I'm not writing it. I'm just telling you then plus summation 3 at a time intersection. I'm not writing, I'm just putting it. So please note, this is the formula that you had already learned dot, dot, dot. This is the formula that you had already learned in your cardinal number property. Remember A union, B union, C, how many number of elements were there? NA plus NB plus NC minus NA intersection B minus NB intersection C minus NC intersection A plus NA intersection B intersection C. So you add one at a time, then subtract two at a time, then add three at a time like that. So that same thing I have scaled up to R sets. Any questions so far? Till this step, anybody has any issue, please highlight. Let me make a partition here also. Okay, now what is NU? How many ways can you distribute the balls where blank groups are allowed, not allowed means no restriction at all. What is the answer? R to the power n, we had already seen that. Minus. Can I say NE1, NE2, etc. till R times will be just R times R minus 1 to the power n. In fact, this R, I will write it as RC1. Why? Because you are picking that one box out of those R boxes which you want to be empty. So that itself can be done in RC1 way. So RC1 and this result I am multiplying, this result which you told me R minus 1 to the power n. In short, I am adding all these. That is the same thing as saying RC1 into R minus 1 to the power n. What about this? How many intersections of two at a time you will get? You will say, sir, there will be RC2 such terms and each one will be, and each one of these expressions will be R minus 2 to the power n. Similarly, the next one will be RC3 R minus 3 to the power n. Yes or no? Minus. It will continue till ever it goes. So this is the formula. I mean, ugly looking expression. I would call it as ugly looking because it doesn't look that good to see. So this expression is the formula for finding the number of ways in which you can distribute n distinct objects into R different boxes or R different groups where bland group is not allowed. Very useful formula. Okay, you will use this formula in your functions chapter also next year to count number of onto functions. See, say to each of these terms is what? R minus 1 to the power n. Let me write it in some different color once again. This is R minus 1 to the power n. This is also R minus 1 to the power n. This is also R minus 1 to the power n. So how many R minus 1 to the power nc would have written? R. What is R? R is RC1 only. Is there any distance between R and RC1? Similarly, this term is what? R minus 2 to the power n. Next term would also have been R minus 2 to the power n. So how many 2 to 2 intersections you can get? That means, in short, how many R minus 2 to the power nc you will get? You will say, sir, RC2 times I will get. So how many combinations are there? RC2. So RC2 times R minus 2 to the power. Same with this. So this is R minus 3 to the power and next one will also be R minus 3 to the power. So how many such terms will be there? RC3 terms such terms will be there. So RC3 times R minus 3 to the power. Got the point? Okay. Now, a good news here is in your school, in your comparative exams, normally the figures are such that you can also solve the question without the use of the formula. That means just by your understanding of division that you have learned a little while ago, you can solve the question. So sir, does this mean this formula is optional for us? To a certain extent, yes, this formula is optional for you. But many a cases I have seen that if you know this formula life becomes easy. So logic, of course, you should be knowing it. Remembering this formula is just a bonus activity. If you happen to remember it, well and good. I will immediately take a question on this before I give you a break. I understand most of you would be eagerly waiting for a break. Just one question I will take on this. We were doing how many different balls? Okay. Yeah, we'll take this question. How many ways, read the question carefully. How many ways can you distribute five different balls into three boxes? By the way, they should have written three different boxes, but anyways, it is understood. And the word here is distributed. Let's not ignore this. The word here is distributed. Not arranged this time. Half the people, they don't read the question only. For distribution, they will do arrangement. For arrangement, they will do distribution. Read the question, what the question is asking. So that no blank groups are there. That means no box should be empty. Very good, Nikhil. See, I'll start calling myself that if I come out, come out with some research papers. I haven't been able to come out with any so far. So I have no right to call myself that. There are people, I mean, of course, getting, you know, finding a novel work in the field of mathematics, which has been an age-old topic. People still find things in engineering, patents are filed and all those things. But finding an age, finding a novel work in mathematics takes a lot of effort. Most of you know already, I'm actually an electrical engineer. I've done my engineering and I'm more close to engineering as compared to, I mean, a research work oriented person in maths. So I happen to know maths because in electrical engineering, we are taught a lot of mathematics. Normal Janta studies for four semesters, but electrical Janta studies it for six semesters. Because while solving circuits, we are using mathematics everywhere, differential equation, Laplace transform, everything is mathematics. DIP was full of matrices. So that's how we happened. That's how I happened to get interested in maths actually. But I'm more of an engineer side of this. So I can apply things rather than doing a research into things. Okay, Karthik, we'll do that. See here, again, if you want to use formula, let's say I'm a person who happens to remember formula, good day for me. And I remember, I know the formula for the second solvent. So the answer, see here, n is 5, r is 3. So if you recall the formula, which I derived, it was r to the power n minus rc1, r minus 1 to the power n, then minus rc2, r minus 2 to the power n, then plus rc3, r minus 3 to the power n, minus, I mean, I'm not writing all the terms. Dot, dot, dot doesn't mean it is going to infinity. Dot, dot, dot means till where r permits it to go. So r is 3, n is 5. So this is 3 to the power 5. This is 3c1, 2 to the power 5. Or let me write it in more, you know, broken up version. This is 3c2, 3 minus 2 to the power 5. Then again, if you see here, 3c3, you have start getting a 0. And of course, you can't even go further, because there's nothing like 3c4, etc. So you stop here. So this is as good as 343. This is 3 into 2 to the power 5, which is 96. And this is 3 into 1, which is 3. So sorry, this is 243, not 343, 243. 343 is 7 to the power 3, not 3 to the power 5. So this is 243 minus 93. Answer comes out to be 150. But let's say I happen to forget the formula. It is quite possible that, you know, see, you're going to write your exams one year, two months from now. How many? 14 months from now, just 14 months. By the way, alarming thing for you, just 14 months from now, you may be writing your first JEMN or first phase of JEMN. So by that 14 months come, you may have forgotten this. This is one of the several things, formulas that will come in your mathematical journey of preparation. So if you, let's say, forget the formula, then how would I do this formula? So without formula, I will show you. Without formula. So without formula, I will remind, I will remember at least that division into groups concept. So I have to divide five group, five distinct objects into three groups with no group blank. That means I can have following possibilities 5, 1, 1, 3, or I can have 5, 1, 2, 2, right? Is there any third possibility? No, right? Correct? Now, if the groups are different, is the order important while you are making these divisions? Yes or no? Order is important or not? Right? Order is important. Correct? So how many ways can you do this first case? Setu will tell me. Setu, tell me the result for this. Unmute yourself if you want to. How many ways can you divide five distinct objects into three groups, two of them having identical number of objects and order is important? Tell me. Tell, tell, tell, tell, tell. Sir, five factorial by three factorial into three factorial. No, one second. Five factorial into one factorial, one factorial. Now here, since two of them are equal, into two factorial and again three factorial and into three factorial. See, so many things you have to account for. Why this two factorial came? Because two groups were identical in size or equal in size. Okay? Why this three factorial came because order is important? Are you getting my point? Or means plus. Then again, five factorial by one factorial, two factorial, two factorial, still one more two factorial. Why? Because these two groups are of equal size. Okay? Into three factorial because order is important. Let's see whether this adds up to give us 150 or not. So first of all, here I will end up getting 120 by two, which is 60. And here I will get 120 by eight into six. 120 by eight is 15. 15 into six is 90. See, same answer 150 ways. So whether with formula or without formula, you can still solve this question if your basics of division is clear in your mind. Is it fine? Any questions? Any concerns? Okay. Here we are going to take a break, quick short break of 15 minutes. Right now, it's this, we'll meet exactly at 6.44 pm after our 15 minutes break. Okay. We'll see, I'll see you with more concepts on the other side. Till then, enjoy your break. Now, this time we'll talk about distribution of or division of whatever you want to call it, division of an identical objects into our different groups. Okay. So here the only difference from the last concept is that the objects now are identical, not distinct. Okay. So we'll be now talking about this. Now, many people say, sir, what about arrangement of n identical objects into our different groups? Now you tell me why I have not talked about arrangement of n identical objects into our different groups. Does it make sense to talk about arrangement of identical objects? Does the ordering of those identical objects really create any difference? Does it really create any difference? See, I talked about arrangement of n distinct objects into our different groups. Then I talked about distribution of n distinct objects into our different groups. But for the case of identical objects, I directly started with distribution of n identical objects into our different groups. Why did not I talk about arrangement of n identical objects into our different groups? Because arrangement of identical objects doesn't make any sense. What will you do if you shuffle the position of these identical objects? Will it create any difference to the arrangement? It is not going to create any difference. So we do not talk about arrangement of n identical objects into our different groups. We only talk about distribution of n identical objects into our different groups. Okay. So let us begin with again two cases. Case one, blank group is allowed. So how many ways can you distribute n identical objects? As you can see, I am not putting any subscript on them. So these are n identical objects. And you want to distribute them into our different groups. Our different groups says that there is no restriction. That means blank groups are allowed. How many ways can you do that? Can somebody tell me? And also let me know the logic behind whatever answer you are getting. Very good Nikhil. Question, problem statement is clear to everybody. You have to distribute n identical objects into our different groups. Very similar to the discussion we had with respect to the arrangement part. How will you do it? Karthik, sorry, I explained the concept. You were not there. You wanted me to briefly explain the derivation once again. Do one thing. Watch the recording several times. That same part. That's the benefit of things being recorded. See, we'll solve this question in two ways. One is consider that since you are dividing it into our different groups, we will consider that there is r minus one identical sticks. So these six are like the partitions which will help you to create groups. Now, two ways to think it. One is by Nikhil's method. Sorry, Nikhil, I am giving that method your name. Nikhil's method, where you are considering that there is n plus r minus one chairs. Why n plus r minus one? Because n identical objects are there and r minus one sticks are there. So on these n plus r minus one chairs, you place your sticks first or you place your objects first. Doesn't matter. So let's say I place my sticks first in r minus one places over here. So for that, I need to choose those r minus one place which can be done in n plus r minus one CR minus one. Now, the remaining place I'm going to put my objects. But is there any scope of shuffling the objects among themselves or is there any scope of shuffling the sticks among themselves? No, because they are identical. So this one n one shows that there's only one way you can place the objects. Just whatever seat is left, you have to put your objects. Nothing else you can do. So your answer is going to become just this expression which is n plus r minus one CR minus one, which is also the same as n plus r minus one CN. Both of them are the same things. So please note this down. This is the total number of ways in which you can place n or you can distribute n identical objects into r places where blank groups are allowed. Not exactly efficient. Your answer will further get divided by r minus one factorial. This is one way to look at it. Another way to look at it is my way which I discussed. So think as if there are n objects. These are n identical objects and there are r minus one identical sticks. Total if you combine, it will become n plus r minus one objects out of which n are identical of type one of one type. And r minus one are identical of another type. Something like the Mississippi equation. So how many words can you make from the letters of the word Mississippi where there is a repetition of let's say four i's and four s's and two p's. So normally what you do, you take the total number of alphabets factorial divided by factorial of the letters which were repeated. Same way here also you'll do. So it will be n plus r minus one factorial by n factorial because n identical objects are there of type one divided by r minus one factorial because r minus one identical objects are there of type two. Sorry for not writing to it. Okay, which clearly becomes again n plus r minus one c r minus one or n plus r minus one c n. Both are same things. Is it fine? Any questions? Any concerns here? Please note this result down. And more importantly, know the logic behind these results. That's going to help you out. All right, now let's take the case where say to see you have these many seats. Seats pay up sticks but how you have these many seats and you have to place r minus one sticks on them. How many ways can you do it? And all the sticks are identical. How would you do it? You'll say, sir, I will choose first r minus one places out of these n plus r minus one place. So that can be done in this many way. And once you place the stick, is there any option of shuffling the sticks among themselves? Let's say stick one, stick two position you changed. Will it create a different way of sitting? No. Why? Identical, correct? Now whatever place is left, what will you do? You'll make the n balls go and sit there. Forget distribution. Distribution is automatically performed. Okay, so let us say there are two sticks like this. Stick one and stick two. Who sat here? Then there's a ball. Then there's a ball and stick. Then there's a ball like that. So you automatically take care of the fact that this is a blank group. This is a blank group. This group will have two items. This will have, will have, you know, depending upon how many items are there like that. Got the point. The second case is where your blank groups are not allowed. Blank groups are not allowed. So how many ways can you distribute n identical objects into r different groups? No group should be empty. Very good, Nikhil. I want others to think. Most of you are not thinking. You're waiting for me to give you the final result. That is not going to help you out. These concepts are not coming from sky. These are all coming from the logic, basic concepts that you have already learned before. You're just applying it. So let's say these are your n identical balls and these are your r minus one identical sticks. You need to place these sticks in such a way that you don't place any number of sticks in front of the first ball or any number of sticks behind the last ball or after the last ball. And you should not put any consecutive sticks in between also. In other words, you're only going to place the sticks in these gaps. How many such gaps are there? n minus one gaps. You have to choose r minus one of them. And that's it. There's no scope of now arrangement of either the sticks nor the balls because they're all identical. So that's it. This is the answer to this question. Is it clear? So please note this is a noun as well. Is it fine? Any questions here? If you have, do let me know. Should we take questions now on this? Okay. Let's take questions. Four boys picked up 30 mangoes. In how many ways can they divide them if all mangoes are identical? Four boys have to divide 30 mangoes among themselves and mangoes are all identical. How many ways can they do the division? Give me an answer in terms of see only. I mean something, see something. No, Nikhil. Think once again. Yes, now it is correct. So is this a case where blank group is allowed or is this the case where blank group is not allowed? Prashiv has all together a different. Listen, why are you doing three factorial? Okay, I understand the first part is wrong, but even the second part is wrong. Why did you do three factorial? Can you tell me the reason for multiplying with three factorial? We just now discussed that if you had n objects and you are dividing it into r different groups and your objects are identical, the number of ways in which you can do it says that blank groups are allowed. So in this case, there could be a boy who could get no mangoes at all. Isn't it? So they have nowhere mentioned in the question that every boy should get at least one mango. They're not mentioned like that. So here there is no restriction. That means you can distribute any number of mangoes, of course, limited to the maximum value to any boy you want. So the answer will be n plus r minus one c r minus one, or you can say n plus r minus one c and both are fine. So n here is 30, r here is four. So the answer will be 30 plus four minus one c four minus one. That is 33 c three. That's it. Okay. Or you can say 33 c 30. Both are same thing. No issues. Is it fine? Any questions? So people who are answering with 29 c three, figure out what mistake happened. Okay, you basically were considering that everybody should get at least one mango, which is not the case. Okay, let's take this question. The way this question has been written is slightly dubious. I'll just make some correction. Find the positive number of solutions. No, find the number of non negative integral solutions. So a small change in the question, please note this down. Find the number of non negative integral solutions of this equation x plus y plus z plus w is equal to 20 under the following conditions. So there are two questions here itself. So first question says zero values are allowed for these variables. Okay. Second question says zero values are not allowed. That means it is excluded. So we'll do first one first. So I would request you to give me your answer by putting the question number one, then give your answer for the first one. Okay, very good, Nikhil. Anybody else? For the first one at least. Not expecting the second one as of now, first one. Is the situation as good as saying I'm distributing 20 identical objects into four different groups and zero values allowed means blank groups are allowed? Do you agree with that first of all or not? If anybody doesn't agree, please let me know. Like people agree with this, isn't it? So think as if there are four children and you're distributing 20 identical chocolates. Such that first guy gets x, second guy gets y, third guy gets z, fourth guy gets w, chocolates. Okay. So altogether there should be 20 chocolates and it could happen that any guy could get no chocolate at all also. Right. So first situation is as good as your blank groups allowed and you're distributing 20 identical objects into four different groups. So n is 20, r is 4. So answer is n plus r minus 1, c, r minus 1, which is 20 plus 4 minus 1, c, 4 minus 1. The answer is 23, c, 3. Absolutely right. Next one, zero excluded means blank group is not allowed. Brand group is not allowed is n minus 1, c, r minus 1. That's obviously 19 minus 1, c, 3 minus 1, oh sorry, 4 minus 1. That's 19, c3. Is it fine? Any questions? Let's take another one. Okay. How many, how many integral solutions are there to this equation? x plus y plus z plus t is equal to 29. Such that x is greater than equal to 1, y is greater than 1, z is greater than equal to 3 and t is greater than equal to 0. Now see, you're doing the same question of distributing 29 identical groups into four different groups. But in this question, you have been provided with a lower limit on the number of items you must give to these groups. That means the first group has to get at least one item. Second group has to get item greater than one. Third group has to get item greater than equal to 3. Fourth group has to get something greater than equal to 0. Okay, greater than equal to 0. So how many ways can you do these divisions or in how many ways or how many integral solutions exist for this equation? Any response? Okay, Nikhil. Okay, let's give a dramatic situation to this particular question. Okay, let's say x is the number of sweets you're giving to, let's say person p1, why is the number of sweets you're giving to p2, z is the number of sweets you're giving to p3, t is the number of sweets you're giving to p4. Okay, now p1 is a person who is coming and saying to you, hey, please give me at least one sweet because maybe I have a child at home who would need that sweet. Okay, so at least one sweet you please give it to me. Similarly, p2 person makes a request, say please give me at least. By the way, greater than one is as good as saying greater than equal to two. Okay, so when you talk about integers and you're saying greater than one, you can only go to from two onwards. So this guy needs two at least. Okay, similarly, the person p3, he needs at least three sweets. Fourth person doesn't have any obligation. You don't give him anything also, he's fine. Okay, so what do you do is you give this person p1 one sweet. Okay, you take, you wanted no one sweet, at least take one sweet. Okay, you gave two sweets to the person two, you gave three seeds to the person three. Okay, let's say you have already given these one, two, three, that is the bare minimum requirement that they have to these students, to these people. So once you have given one sweet to person one, two sweets to person two, and three sweets to person three, you are already given six sweets or six identical chocolates. So ultimately you are left with 23 identical chocolates only. And now what you're doing, these 23 identical chocolates, you are distributing to these four people without any restriction. So let's say the first guy gets capital X, second guy gets capital Y, I'm not changing the variable here. Okay, so instead of small x, small y, I'm putting capital X, capital Y, just to keep those two equations different. So now here you are trying to distribute the sweets such that bland groups are allowed, bland groups are allowed. So how many ways can you do this? So n is 23, r is 4. So as per the formula, it is n plus r minus 1, c r minus 1. So answer is 26, c 3. Is this fine? Any questions, any concerns? Any questions, any concerns? Is it fine? If you have any questions, let me know. Okay. So these kinds of finding the number of non-negative integral solution type of question is very, very commonly asked. And it is not only a type of question asked in J, I've seen these kinds of questions asked in regional entrance exams also. Okay. So please get this idea very, very clear. Okay. Should I give you a question now? Okay. Find the number of non-negative integral solutions to this inequality. x plus y plus z is less than equal to let's say 30. x plus y plus z less than equal to 30. Now instead of an equation, I gave you an inequality. Find the non-negative integral solutions to this particular question. Just meaning you are listing half a hazard. You already gave one to the first guy, two to the second and three to the third. Six weeks are gone. No. Because the first guy had a requirement of x greater than equal to one. Second guy had a requirement greater than equal to two. Third guy had a requirement greater than equal to three. So minimum requirement is one, two, three. So you gave it to them first of all. In the beginning, only you gave it. There's only one way to do it. Take one suite, give it to that guy. It doesn't matter which one you are taking. All are identical. So that remaining six weeks that you're talking about, that is not remaining, that was already taken care before we started the distribution of the 23 objects, 23 identical suites. Say to your question, I did not understand. Better you phrase it in a proper way or you speak out. That is the total number of solutions possible for x, y, z as a combination such that they satisfy that equation. For example, one answer could be 5, 10, 4. Sorry, 14. 5, 10, 14. That is one combination. 15, 4, 10. That is another combination. Got the point. So all those combinations are taken care in that 23 C3 or whatever we got, 26 C3 combination. 4 to the power 30. Good. Can you justify your answer Prasoon? Why 4 to the power 30? Why not 7 to the power 30? Why not 100 to the power 30? Some random answers you are giving since morning, since afternoon today. Okay. Nikhil, very good. Okay. Anybody else? Okay. See, this question can be solved in two ways. The first approach is, I would say not so smart approach where you start making cases. See, if you have lesser than equal to 30, you can start with 30. You find the number of solutions for this. You find the number of solutions for this. You find the number of solutions for this and go all the way till you find the number of solutions for this. So whatever solutions you get for these, you start adding them. Okay. By the way, how many solutions are there for this? 32 C2. Am I right? How many solutions are there for this? How many solutions are there for this? 31 C2. This will be 30 C2. And can I say this will go all the way till last term, which is 2 C2. That is only one solution where everything becomes 0, 0, 0. Okay. So now you need to sit up and add. So your total number of ways would be the sum of 2 C2, 3 C2, 4 C2, all the way till 32 C2. How do you add them? Anybody knows the way to add them? Anybody knows the way to add them? Okay. So here is a method to add them. See, this 2 C2 is anyways 1. Can I write it as 3 C3 also? Because 3 C3 is also 1. Okay. Now see the beauty. A very interesting thing you will see here. If you recall, there was something called the Pascal's identity. What was Pascal's identity? NCR, NCR minus 1 is equal to N plus 1 CR. Correct? If you recall, we had done this. This is called the Pascal's identity. So what is 3 C3 plus 3 C2 as per Pascal's identity? Tell me, tell me, tell me. Write it down on the chat box. 4 C3, very good. What is 4 C3 plus 4 C2? 5 C3? What is 5 C3 plus 5 C2? 6 C3? Can I say continuing so on, you will be finally left with 32 C3 plus 32 C2 and that will finally give you 33 C3. But this approach is not a smart way to solve the problem. But even if you want to take this approach, I mean, you should be aware of this. By the way, this is something which many people are not aware of. RCR, R plus 1 CR, R plus 2 CR, all the way till, let's say NCR, this is N plus 1 CR plus 1. This is an identity which we normally call it as hockey stick identity. Why this called hockey stick identity? See, it comes from the, it comes from, let's say the Pascal's triangle. Let's say I draw a Pascal's triangle. Sorry, 6. So here if you see RCR, R plus 1 CR, etc. So let's say I choose some value. Let me take these values themselves. Let me take R as 2. And let's say I start from 2 C2, 3 C2, 4 C2. That's it. And up here. Now see what is going to happen. 2 C2 is this term. This is 2 C2. Okay, 3 C2 is this term. Okay, 4 C2 is this term. Now if you see the next row that I'm going to make, this is the next row, these three terms add up to give you this value, this guy. And as you can see the way I've drawn this arrow, it looks like a hockey stick. Okay, that's why this term is actually 5 C3. Okay, so 2 C2, 3 C2, 4 C2 is 5 C3. Okay, so this is called the hockey stick identity. You can apply it to any, any file. Let's say you want to do these number, then this will add up to give you this 10. You can check it out. Okay, you can take a smaller one also. Let's say these two will add up to give you a 2. Okay, these two will add up to give you a 5. So as you can see, it's like a, bent like a hockey stick. I hope you have all seen a hockey stick. Okay, anyways, so this is our identity, which you can note down. This is called the hockey stick identity. Please note it. Now this is again, as I told you, not a smart way to solve the problem. So I'll show you a smarter way to solve the question. The smarter way to solve this question is by the use of something which we call as a dummy variable. So using a dummy variable, use a dummy variable. What is a dummy variable? As you know, dummy, dummies, that doesn't exist. It has been made up. So what do you do? You take a dummy variable W or D. Let's say says that the dummy variable makes this inequality as a equality. So for that, this dummy variable should be greater than equal to zero. Am I right? Isn't it? So if this guy is less than 30 and you want to make it 30, you have to add some quantity which is greater than equal to zero. Isn't it? And you also wanted a non negative integral solution, which means even your XYZ should be greater than equal to zero. Now isn't this a case where you are solving a distribution of 30 identical objects into four different groups where blank group is allowed? So what's the answer? n plus r minus 1 C r minus 1. Straight away, you get 33 C 3, which was the answer you got here as well. Right? So see how fast is the use of this dummy variable logic or concept to solve this question? Is it fine? Any question in either of the two ways? Let me know. Let me know anything anywhere you want to ask me. Nikhil, how did you solve this question? Dummy variable or the long away? The long away? Is it fine? Any question, any concerns? Okay, so we'll take one more question then. Are we already for the next question? Okay. So next question is, let's take this question. Question is, if three dice are rolled, in how many ways can the sum of the number of the numbers appearing on the top face of the die be greater than 7? In how many ways can some of the number appearing on the top face of the die be greater than 7? Okay, so D1, D2, D3, let's say these are the top numbers appearing on the top face of the die. In how many ways can this be greater than 7? Okay, let me just okay, let's see whether you're able to solve this because this question is slightly complicated. Keep it 6. Sorry, if I'm changing the question in the middle of your solution, I'm so sorry about it. Keep it 6 to be on the safer side. In how many ways can the sum of the numbers appearing on the dice be greater than 6? So last question of the day, I'm expecting people to answer this. Okay, Sethu. Dice, distinct dice, three distinct dice, three different dice, D1, D2, D3. The sum of the numbers adding, the sum of the numbers coming on the dice should be greater than 6. Three distinct, of course, three dice means three distinct dice only. It's like saying three distinct people are coming. Three people is there, automatically three distinct people. Sharduli finally woke up after her sleep. Okay, Sharduli. Anybody else? So this question is a simple thing. I mean, here first of all, can I say the number of ways to do it is total possibilities minus total possibilities minus those cases where D1 plus D2 plus D3 is less than equal to 5. Correct? Now, many times we can add, you know, find this also manually. Right? But let's do it by using our dummy variable concept. So let's say D1 plus D2 plus D3. Let's introduce a dummy variable W equal to 5. So in how many ways can you distribute these five distinct objects into this? Now, W should be greater than equal to zero. Okay? And here, please note one very important restriction that many people would forget in this question is that D1, D2, D3 should be greater than equal to 1. In the dice, there is no zero. Minimum is 1. Okay. Right? So think as if you have to give 111 to each one of these guys. So let's say D1 dash D2 dash D3 dash plus W will become 2. Right? So what do you have done? You have given 112 D1 D2 D3 and you're left with only D dash. Another way to look at it is let me write it in another way, which is more easy to understand. Let's write D1 as D1 dash plus 1. Let's write D2 as D2 dash plus 1. And let's write D3 as D3 dash plus 1. Okay? Now here, please understand that as D1 is greater than equal to 1, your D1 dash will be greater than equal to zero. As your D2 is greater than equal to 1, your D2 dash will be greater than equal to zero. And as your D3 is greater than 1, D3 dash will be greater than equal to zero. So substitute these in place of D1, D2, D3 over here. Okay? So when I do that, I'll end up getting D1 dash plus 1, D2 dash plus 1, D3 dash plus 1 plus W is equal to 5. In short, you will have D1 dash, D2 dash, D3 dash plus W is equal to 2. So how many ways now you can distribute it without any restriction on any one of these groups? You'll say 2 plus 4 minus 1 C4 minus 1. Am I right? That is nothing but 5 C3, which is 10. So out of total possibilities, you need to subtract 10. Total possibility is 6 cube. And out of that, if you need to subtract 10, your answer will be 206 ways. Is it fine? Any questions? Any concerns? Oh, I said greater than equal to 6. No, greater than equal to 6 means 7 onwards, right? Oh, 7 onwards means less than equal to 6. Oh, oh, Rama. Sorry, my mistake. So this is going to be 6. This is going to be 6. So this is going to be 6. This is going to be 3. So this is going to be 3 plus 4 minus 1 C4 minus 1, which is going to be 6 C3. 6 C3 is 20. So subtract 20 from 200 and oh, I'm so sorry. Subtract 20 from 216. That's going to be 196. 196 ways. From where did I get 216? Seethu, you tell me. If you roll 3, how many possible combinations of XYZ you will get or D1, D2, D3 you'll get? 6 into 6 into 6? 216? Correct? No. Yes or no? Is it fine? No. There is some problem. There are some issues which this approach will not be able to address. And those issues are when there is an upper cap on the values of the variables. So that is something which we are going to deal in our next class. But our next class will be a shorter one on PNC. We'll be able to cover it in I think the first one and a half hours. The remaining part of the class, which topic do you want me to start with?