Hello all welcome to the YouTube live session on permutation and combination by symptom Academy So guys those who have already joined in the session I would request you all to type your names in the chat box so that I know who all are attending this session we'll be starting the sessions sharply at 1 p.m. and Tentatively it would go for three hours and if the response is there We can also extend the session by half an hour 45 minutes more So again Request you all to type in your names in the chat box so that I know who all are attending this session All right, so hello also, let's start the session now. I think almost everybody has joined in Do let me know if there's any problem with the audio or the video quality so that I can get it rectified as soon as possible Okay, so I could see a lot of people advaith ved abhay barat shiram ashtosh Lakshya sushant andrew khyati Hello, good afternoon everybody. So let us start with the session of permutation combination where we left off in the classroom session So today's class I would be starting with the ways of distribution. So the concept which I'm going to start with is distribution of distribution of identical things and identical things into our different groups into our different groups So just to be very clear on what we did in the class in the class. We talked about division of n distinct things into groups We talked about arrangement of n distinct things into our different groups We also talked about distribution of n distinct things into our different groups Where groups can be blank or groups can not be blank So today's session I am going to start with distribution of n identical things. This is very important. We are talking about identical things into our different groups Hi, umkar. Hi, uh, samir sukirt So we just started with the session So my first concept would be for today the distribution of n identical things into our different groups Okay, so here also we'll start with two things one Where blank groups are allowed Okay blank groups Are allowed So first we'll talk about those case where you're trying to distribute n identical things says that Some of the groups can be blank or some of the groups can be empty Okay Now I would like you all to first think about it for half a minute or so and then we'll discuss the concept Think as if there are let's say n identical let's say laddoos Okay so there are n identical laddoos And let's say I have to distribute these laddoos to our children Right, so our children are there And you want to distribute these n identical laddoos through these our children and unfortunately Some students may not or some children may not get any laddoos at all. That is what I mean to say blank groups are allowed How will I do this task? How will I accomplish this task? Just think about it for a minute or so And do let me know through your chat box What comes in your mind when you want to accomplish this task anybody any any suggestion that you have so there is no restriction whatsoever Anybody can get any number of laddoos provided I mean he cannot get more than the maximum number of laddoos and there can be situations where A child goes without any laddo Okay, now if it is too big to type, let me just explain this now Try to imagine this situation as if you have these n identical objects or n identical laddoos and You are trying to distribute it between our people So to distribute between our people you would require r minus 1 partitions Right If you take r minus 1 partitions You can call it as partitions or you can call it as sticks right And if you start placing these partitions in between these laddoos Then what these partitions will do these partitions will automatically start No separating out these objects into different different groups for example, I start with Two sticks over here. Let's say one stick over here One stick over here like that. I started putting my sticks Okay, what does it mean? It means that the first child has got no laddoos because there's nothing over here, right? So this is cross that means he gets no laddoos or zero laddoos Even this is this doesn't have any laddo in between. So second child also ends up getting no laddo at all Whereas the third child will now get two laddoos Right, the fourth child will get three laddoos and like that it will start Segregating the laddoos between the our children Is this approach clear in your mind? So what I have done is I have started using sticks and how many such sticks r minus 1 sticks So these r minus 1 sticks will start segregating these laddo out Among these are children Now guys, how do you do this? How do you find the number of ways of placing these sticks? between these laddoos now think as if think as if you have n objects of type 1 And r minus 1 objects of type 2 Right, what is this type 1 object type 1 object is your laddoos What is your type 2 object type 2 object is your sticks, correct? So what are you trying to do you're trying to arrange? Arrange these objects among themselves Right, so you have n objects of type 1 r minus 1 objects of type 2 And you are trying to arrange them among themselves Yes, Sukirti your answer is correct So how do I do that? It's like the same situation where you have a word like let's say Mama Okay How many different words four letter words can you make from mama? You would say your answer is going to be four factorial by two factorial two factorial correct So do you have two letters of type one? And two letters of type two and the number of ways you which in which you can arrange them Is nothing but total number of letters factorial divided by factorial of the number of repeated objects In the same way the situation is here you have all together n plus r minus 1 objects, isn't it? n objects of type 1 and r minus 1 objects of type 2 So total you have n plus r minus 1 objects Out of which you have n repetitions of type 1 And r minus 1 repetitions of type 2 So your answer is going to become In a similar way as I did this problem mama your answer is going to become n plus r minus 1 factorial by n factorial n minus r factorial Right, so this is playing the role of this four And these two are playing the role of these two factorial through factorial right which Sukirt has actually written it as n plus r minus 1 c r minus 1 That's the right way to express it because we know from the formula of ncr That I could express this as this term However, you can also write it as n plus r minus 1 c n So both these answers are going to be the number of ways in which you can distribute n identical objects into r different groups So both are correct answers request you all to make a note of it This is a very very Important concept in permutation combination soon. We'll be taking a lot of problems based on that Okay Now the very same concept The very same concept can we address this by using multinomial theorem? Let's try to understand this. Can I get the same answer? These two answers can it be obtained by the use of multinomial theorem? I think last class we had started talking about multinomial theorem Where we talked about distribution of distinct objects into r different groups Right, so now I'm going to make use of that concept yet again So I would request all of you to be very very attentive Yes, Bharat.  Yes, Bharat that you're absolutely correct. There's another way of thinking about it Now I'm going to approach the same problem through the use of multinomial theorem Okay, so let's talk about how we can do it through multinomial theorem multinomial theorem Now in use of multinomial theorem friends, what do we want over here? We want to Uh distribute n objects into r students, right or n identical objects into r students So there is a chance that the first student can get zero object One object Two object or you can get all the n objects, right So this is about your first student getting the number of objects So these powers signify how many objects that person that is the first student can actually get He can get zero one two all the way till n In a similar way the second student can also get zero one two All the way till n And this may continue for all the students. So basically till the rth student That person can get zero one two Till all the n objects, right Now what are we trying to find out in this multiplication of these multinomial terms? We are trying to find out the coefficient of x to the power n in this Because ultimately you have only n objects to distribute So what I'm trying to do is I'm trying to see in how many ways Can I achieve x to the power n term from this multiplication? Which indirectly tells you How many times you can generate x to the power n means it is Trying to tell you how many ways you can distribute your n objects Between or among those r students Right to give you an example. Let's say Your n is 10 Okay, your 10 x to the power 10 coefficient This coefficient will keep a track of how many times x to the power 10 has appeared in the multiplication of these terms So it can appear in various ways like You know, you can have a one from here. You know, let's say there are only four children. So one from here Let's say you pick up two from here. Let's say you pick up three from the third term And let's say you pick up four from the fourth term So this is one way of generating x to the power 10 Another way could be you pick up three from here. You pick up three from here You pick up one from here. You pick up again three from here. That is another way to generate x to the power 10 So who will keep a track of how many times x to the power 10 term is being generated? This coefficient will keep a track of that. So coefficient is like A counter it counts Right, it counts. How many times x to the power 10 term is going to be generated No, see we don't need a factorial Bharat because you are not trying to arrange it Because objects are identical Right, so the expression is slightly different in this case Try to understand here We are not trying to shuffle the objects among one another because here the objects are identical Factorial terms will appear when your objects are distinct from each other Is that clear Bharat? Right people who have questions, please feel free to type in in the chat box All right, okay, cool. So now One very important thing that I wanted to point out is If you look at these kind of terms, they are all in GP Right, this is a GP, right? So if I take the first term that is x to the power zero x to the power one x to the power two I think in the class itself I had to explain you this is a geometric progression Okay, this is a geometric progression correct In a geometric progression You basically can find the sum of it by using the formula sum of n terms is a One minus r to the power n by one minus r. I think we had discussed this the other day Where a is your first term and r is your common ratio R is your common ratio Correct, so in this the first term is going to be x to the power zero which is nothing but one Okay, and common ratio is also x And altogether there would be n plus one terms Right from zero to n there would be n plus one terms divided by one minus x Correct, so here your a is x to the power zero which is one r is your x Okay, now guys one thing you will note here that You need coefficient of x to the power n But you already got x to the power n plus one over here So when you're writing this term you would realize that You are writing this term Repetitiously r number of times Okay, you are writing this r number of times Okay, so you're writing this r times And you're trying to find the coefficient of x to the power n in this term Now you tell me would this term contribute To generating x to the power n Type in yes or no on your chat box Do you think x to the power n term is going to contribute towards generating an x to the power n term? In fact all these terms will they contribute towards generation of x to the power n Just type in yes or no in your chat box I'm waiting for your response guys Sushant, why do you think it will it'll contribute towards x to the power n it is already higher in power than x to the power n It is already higher in power Than x to the power n right So such terms will never contribute towards generating x to the power n right Yeah, otherwise my question is If you want to create an x to the power n term, let's say I want to create x to the power five term Would a term like x to the power six et cetera will will that contribute towards generating x to the power five This term will never contribute towards the generation of x to the power five because it is already higher in power than x to the power five Okay, yes, I'll talk about it. Sukrit. I'll talk about the denominator as well But in the numerator such terms will never contribute towards generation of x to the power five So my point here is that Let me go to the next slide My point is that when you are finding the coefficient of x to the power n in One minus x to the power n plus one by one minus x whole to the power r This term can be safely neglected This term can be safely Neglected that means it is as good as finding coefficient of x to the power n in One by one minus x to the power r No, it is not about extraction. Shiram. We don't want to extract any term from it We want to see whether x to the power n is getting generated Wholes only not being extracted from certain terms Okay, so guys the point where I want to why did I want to make over here is that if if The maximum number of objects That a group can get group can get Is equal to the total objects present Total objects present with you See, what do I mean? Maximum number that a child can get he can get all the n laddus, right? And that is also equal to the total object present Then instead of writing like x to the power zero x to the power one till x to the power n Whole race to the power r You can write this as x to the power zero x to the power one all the way till infinity also Are you getting this point? So it makes no difference if your number maximum number of objects a group can get Is equal to the total objects present Okay, so we know some till infinity in a geometric progression Is given by a by one minus r That is in this case. It should be x to the power zero which is this term By one minus r which is actually one by one minus x So do you see this term one minus one by one minus x has been getting used over here However, this is just an observation which you can follow to save your time Even if you decide to take it till n terms your answer will not go wrong Okay, so this is just an observation which is which can be a shortcut for you in order to save your time You know, you don't want to write this term. That is fine You can directly write the sum till infinitely many terms And that will do the same job for you all right Now why it becomes easy is because now it has changed to finding the coefficient of x to the power n in one minus x to the power of Minus r Yeah, I mean it's fine x doesn't have any meaning over here Bharat So it you are correct x should be less than one else this series will not be a convergent one But x does not does not have any significance over here. It is just meant for addition of powers The role of x is just to facilitate adding of powers It has no functional uses Is that okay Bharat? Yes, that's what Sukirti is also saying Yeah, r is different from n r is the number of people to which you are distributing n is the number of objects Then how come numerator and denominator have x? See, uh, Sushant you are getting confused over here The role of x is just to facilitate addition of powers Addition of powers if it becomes n Then the coefficient of x to the power n gives you how many ways you can generate that power How many ways you can generate that power means how many ways you can make that distribution? So x doesn't have any functional role in this case x only helps in addition of powers Is that clear it has nothing to do with r different from n or n different from r See if you go back to the previous slide Let me just show you the previous slide Uh, what is happening in this slide? in this slide In this slide you are trying to raise x to different different powers.  Now when this x multiplies with this x it actually leads to addition of powers So in order to find x to the power n I would have Yeah, now it should be proper. All right, so let us resume from where we left off So basically all I need to do at the end of this exercise is get the coefficient of x to the power n From 1 minus x whole to the power of minus r Okay Now how do I do that? For that I have to just learn a little bit of binomial expansion.  There will be infinitely many terms in this Of course as Bharath says mod x should be less than 1 in this case, but right now I'm not going into details of this expansion I'm just uprising you of the The way to expand this binomial term x plus 1 to the power r. This is not x plus 1. This is 1 minus x Now if you look at this term very very carefully, you would realize that Minus r into minus x will create an rx Correct in the second term also The minus of this and the minus of this will cancel each other out and you get r plus 1 r into r plus 1 by 2 factorial x square Okay And here also we have r r plus 1 r plus 2 by 3 factorial into x cube So what is happening here is that three negative powers are coming from here and one negative power will come from here So they will all cancel the negative powers out So you would realize that none of the terms would have a negative power in it negative sign in it Okay, so this is plus rx. This is plus r r plus 1 by 2 factorial and so on and so forth Okay, now If I want to go till some x to the power n-th term, let's say I keep on going and I reach some x to the power n-th term What is going to be the coefficient of that term? You would say pretty simple. It would be r r plus 1 r plus 2 all the way till you reach r plus n minus 1 Whole divided by n factorial, isn't it? If you see that trend for example, let's say x x to the power 3 It started from r r plus 1 r plus 2, right? So it ended 1 less than 3 over here. So this 2 is 1 less than 3 Correct So the same way if I'm going till n terms It will go till r plus n minus 1. Is this part clear? And in the denominator, you have a 3 factorial if there is a 3 over here You have a 3 factorial over here. So if I have a n over here, I would have a n factorial over here Is that clear? Now, if you look at this coefficient Coefficient of x to the power n It is actually something which looks very ugly Right now it's in a very early shape There are so many terms involved But I can always make it super simple How?   You can write that as n plus r minus 1 c n or n plus r minus 1 c r minus 1 Did you see guys? I ended up getting the same answer I ended up getting the same answer But now by a very very different approach called the Binault multinomial theorem Okay Now one would think Why will I use multinomial theorem when I could do without it very easily? Please let me tell you when situations will become tough like When I put a restriction of the upper and the lower limit for example, if I put a restriction that Uh, nobody can get less than two laddus and nobody should get more than eight laddus When such restrictions are imposed in the problem You will have to use multinomial theorem to solve that problem Otherwise if there is no restriction, you don't need multinomial theorem Asushant it is not merely an alternate method. It sometimes becomes the only method to solve the problem Okay, you will soon realize when I start solving questions based on this concept right So guys, uh, let me finish off this case by uh taking the second case where blank groups are not allowed So so far we were talking about blank groups are allowed Now, let's take the case be where blank groups Are Not allowed So in how many ways can you distribute? n laddus Again, let's get back to laddus. Hope you all had your lunch and all Okay, so Nobody is feeling hungry listening to laddus Yes, yes, we will solve this by using multinomial as well Okay, now you you have r minus one n laddus and r minus one sticks Okay, now tell me how many ways can you Arrange these n laddus and r minus one sticks so that No stick can come No stick can come in these two positions Okay, Varad, let's use chocolates then Yes, guys answer, please I think you know the concept well enough now to give me the answer so you have um r minus one sticks and you have n identical objects and you want to place these r minus one sticks In between the gaps of these n identical objects So that these cross positions you cannot put any one of the sticks Exactly Sukirti were absolutely correct Sushant is also correct So it's very obvious between between These n objects we will have n minus one spaces or n minus one gaps Yes or no So if you count these gaps Okay, you realize that there would be n minus one gaps Out of n minus one gaps You have to choose r minus one gaps to place Your r minus one sticks Which you can easily do by n minus one cr minus one. So this becomes your answer Right, isn't it simple So if you realize it through your analogous situation, you would realize things become very easy to understand and manage Okay Now I would request all of you to use multinomial theorem Okay Let's use multinomial theorem for this all of you Just open up your notebooks Just try to see whether you can get the same thing through multinomial theorem. I'll help you out also so in multinomial theorem We'll have to search out for coefficient of x to the power n in Now, what is the minimum number of objects that a person can get if nobody should go without any object? Right, Sukirt, just remove x to the power zero terms. So we'll start with x to the power one And as I told you you can go to infinity because your maximum number of objects is equal to the Total number of objects you can distribute Okay Right, so Bharat you are talking about the next step here. So it is about finding the coefficient of x to the power n in Take x to the power out common So it'll become one plus x all the way till infinity to the power r Okay, which you can again rephrase as saying coefficient of x to the power n minus r in One plus x x square all the way till infinity whole to the power of r So it is like saying One by one minus x to the power r. We are finding the coefficient of n x to the power n minus r in this Now we did this exercise a little while ago. So I I hope from there you can continue and give me the answer All of you just write out on your notebook So it's finding coefficient of x to the power n minus r in One minus x to the power minus r because Sushant we dropped the factor of x to the power r from here This is dropped. This is dropped from both the sides See, uh, if if I say find the coefficient of x to the power five take an example Let's say x to the power five in x square one minus x to the power. Let's say minus four Okay, what will you say? If I'm finding the coefficient of x to the power five in this It is as good as saying or it is equivalent to saying I want x to the power three in this term, right? So I dropped x to the power two from both the sides and made it three Is it clear Sushant now is this example clear enough to you why I why I took n minus r Okay Yes, so basically the answer that you'll still get over here is n minus one cr minus one Which is the very same answer which I got for the previous case So whether you use multinomial or whether you don't want to use multinomial in this case Only in this case the answer comes out to be the same However, as I repeated multinomial theorem would be useful in many situations Which we will realize today while doing some problems So just to recap what I did so far today recap is If you want to distribute n identical objects Into r different groups Are different groups such that blank groups are allowed blank groups are allowed Then the number of ways to do that will be n plus r minus one c r minus one or n plus r minus one c r Okay, and if you want your blank groups to be not allowed blank groups Are not allowed Then the number of ways to do that will be n minus one cr minus one So this is the crux of the story so far. Let us now solve Problems based on that Uh, Bhanath I didn't get your question. We extend it to n minus one right means Guys, let us take questions now The first question is In how many ways just a very basic question in how many ways in how many ways Five identical balls Can be distributed Can be distributed into three different boxes Into three different boxes So that no box remains empty Yes Shushant you are correct Sukirt is also correct So it's very simple. It is just going to follow a simple one line solution What is that? n minus one cr minus one So it's five minus one c three minus one Which is going to be four c two four c two is going to be six Okay Any alternative way to solve it? If I don't want to use my formula, can I solve it in an alternative way? Yes, Shri Ramadwath. You're all correct Ashutosh Let's look an alternative way to solve it So, uh Alternately, let's say we have these three boxes Okay box one box two box three Correct So what we can do is We can have the following ways in which we can put the balls. We can have one One three Correct, right And we can have one two two Of course, there can be inter shuffling of these, uh, you know, uh numbers for example, I can have Three one one also one three one also All those things Now in how many ways can I give one ball to box number one? One ball to box number two And three balls to box number three No, I'm not using multinomial theorem. I'm using my common sense way to solve it So how many ways can I give one ball to box number one? Please type in your answer in the chat box You have five identical balls in how many ways can I give one ball to box number one? Exactly, it is only one way right in the same way. I can give one ball to box number two also in one way And three balls to box number three also in one way, right? So it is just going to be one And these numbers can be intra shuffled among themselves in three factorial by two factorial ways It is like saying how many Uh, three digit numbers can you make from hundred and thirteen? So you'd say hundred and thirteen one thirty one Three one one right so only three possibilities So that is what is given by this term So which is going to be three Correct in the same way you can distribute Your five different identical balls to these three different boxes in one way And again in this you can do the shuffling in three factorial by two factorial ways, which is again going to be three So these two are cases for me case one case two So my answer will be the sum of these two by fundamental principle of addition. I am still going to get six So see I'm giving you so many ways to think about it So here my a number of balls were very very less hence I could manage it with this But when things become complicated Of course, uh, you know multinomial theorem or these formulas will be very very helpful in that regard that clear Is this problem clear any question? Please feel free to type it in the chat box Write clr if it is clear to you Okay, so moving on to the next question This question is how many ways in how many ways four boys distribute 30 identical mangoes 30 identical mangoes among themselves such that each boy gets at least two mangoes Great great to see that it was clear to you now. Please attempt this next problem Read this question very carefully You four boys want to distribute 30 identical mangoes among themselves such that each boy gets at least two mangoes Take your time work it out. I think you want to do shovel In fact, I would suggest that if you already have a ready-made formula go for it 2 keys is 25 c3. Okay Let's check it out. It's fine. Don't don't calculate your answer Just write it in 25 c3 26 c3. Whatever you are getting so shiram says 28 c2 Advait also thinks the same 28 c2. What about others Shushan's 25 c3. Okay, so two people say 25 c3 two people say 28 c2 Guys and girls. I want everybody to participate. So please feel free to type in your answers Okay, Ashutosh shaman also say the same thing All right, so guys, okay. Bharat also says the same thing. So mostly you people are saying 25 c3. Let's check it out so guys, let's say X1 is the number of mangoes the first boy gets. Okay, so let Let boy one get x1 mangoes boy two get x2 mangoes boy three gets x3 and boy four gets x4 mangoes Okay, so first thing first If you add all the mangoes it should come out to be 30 Okay, and not only that All these values whether you're saying x1 x2 x3 x4 They should all be greater than equal to two Now here we have put a restriction Now see how do we work around this with this restriction? What I would do now, I would Treat x1 as some other variable x capital x1 plus two and I say my x1 should be My x1 should be Greater than equal to zero Right, this is equivalent to saying that your small x1 is greater than equal to two So what I did I wrote small x1 as capital x1 plus two And said capital x1 should be greater than equal to zero Which automatically would imply that small x1 would be greater than equal to two Is this clear in the same way I can say let x2 be capital x2 plus two So if I say capital x2 is greater than equal to zero that automatically implies x2 is greater than equal to two Right Bharat, I'm trying to do the same thing but in a more mathematical way Why I'm trying to do this way is because later on I will give you another question Right where this thing will be very helpful to you. So I'll solve it in multiple ways So in the same way I would do this And say x3 is greater than equal to zero which automatically implies x3 will be greater than two, okay And in a similar way I would take x4 as capital x4 plus two and I say If capital x4 is greater than equal to zero my small x4 will be greater than two So what I do I make these replacements I replace it in this equation That means this entire thing will become x1 plus two plus x2 plus two plus x3 plus two plus x4 plus two Equal to 30 Which means x1 plus x2 Plus x3 plus x4 will be equal to 30 minus 8 which is 22 Now remember here that your x i capital x i has no restriction. It should be just greater than equal to zero Right So think your n now is 22 r now is four And blank groups are allowed blank groups are allowed So let's recall the formula. What was the formula we learned? n plus r minus 1 c r minus 1 right Or you can say n plus r minus 1 c n both are the same things So my answer will be 22 plus 4 minus 1 c 4 minus 1 Which gives you 25 c 3 or 25 c 22 Both are correct answers for this question now This is a mathematical way of looking at it What you can do is what Bharat said you can first distribute Two two mangoes to everybody Which you can do it in one way only right So another way of doing it is first Distribute two mangoes to each two mangoes to each The number of ways of doing it is one The number of ways of doing this is only one way correct And second is if you're left with 22 mangoes And you are distributing it to four people or four boys You can directly use this formula which I discussed with you right So it's one and the same thing But the method which I told you would be helpful to you in other cases as well Right. So Bharat again All these formulas have been can also be derived by using multinomial. So use multinomial theorem very judiciously Right. Do not misuse it because it comes with a cost of time Okay Let's go on to the next problem next question is find the number of non negative Integral solutions of This equation x plus y plus z plus w equal to 20 Okay, so I'll repeat the question first find the number of non negative integral solution Of this equation x plus y plus z Plus w equal to 20 non negative integral solution means This should be greater than equal to 0 Okay, so if you have understood the concept it becomes so easy. It is just like Trying to distribute 20 mangoes to four boys Such that blank groups are allowed Yes or no such that Blank groups are allowed It makes it so easy.  Yes Other y 12 c 3 Was that was that a typo error? Let's move on to the next question The number of integral solutions The equation x plus y plus z plus t is equal to 29 when 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 12 c 3. Why it would be 12 c 3 No, it will not be 22 c 3 because the formula is n plus r minus 1 c r minus 1 or n plus r minus 1 c n r is the number of groups And n is the number of identical objects So when you're trying to distribute 20 between x y z and w, it is like saying you are distributing 20 identical objects to four people So this would be 20. This will be 4 minus 1. So it will become 23 So 26 c 3 is absolutely correct. Sukirt How it is clear to you why not 22 c 3 why it was 23 c 3 Can we solve this problem now? I hope everybody has tried it out. I could just see two people answering it Bharat also says 26 c 3 good. Where is 10 identical objects coming from? They're 20 identical objects Right, let us solve this problem Now guys here again, we'll follow the same approach since your x should be greater than equal to 1. What I'm doing is I'm taking x as capital x plus 1 And saying that my capital x can be greater than equal to 0 Okay Since now y is greater than 1 it implies y should be greater than equal to 2 So what I'll do is I'll take y as capital y plus 2 and say This is greater than equal to 0. Okay Since it is greater than equal to 3 I can say z is equal to capital z plus 3 And capital z is greater than equal to 0 Right and t will remain t So make these substitutions Here So I'll get x plus 1 plus y plus 2 z plus 3 plus equal to 29 Which means x plus y plus z plus t Is equal to 29 minus 6 which is going to be 23 Okay Now your n is 23 r And blank groups are allowed blank groups are allowed So basically this is done just to allow blank groups So blank groups Are allowed when you do this So My formula is pretty much there for me. I'll use my formula n plus r minus 1 c r minus 1 which means 23 plus 4 minus 1 c 4 minus 1 which is 26 26 c 3 Or 26 c 23 both are correct So whoever has got it congratulations, you're correct Guys now what I'm going to do is I'm going to complicate the scenario a bit. Okay. So so far whatever you have got Was very much a straightforward application of I know the formulas that we had learned so I'm going to complicate things a bit All right So next problem all of you please Read the question very carefully and try to solve it Next question is how many how many Integral solutions How many integral solutions? are There to the system are there To the system x 1 plus x 2 plus x 3 plus x 4 plus x 5 is equal to 20 x 1 plus x 2 is equal to 15 when when when All your x size that is x 1 x 2 x 3 x 4 x 5 are greater than equal to zero Okay So think very carefully take one minute or so and then give me the answer for it. No need to rush Is it 20 no Sukirt the answer is more than that Just think of it. It's very easy question. You just have to think it out. Just take some time No sort 7 c 2 shushant shawan. Sorry Again, I'll repeat the question. How many integral solutions are there to the system x 1 plus x 2 plus x 3 plus x 4 equal to 20 Okay, I think you're asking about this 20 and let me write it properly This is and your x 1 plus x 2 should be 15 16 times 7 c 2 Oh, by the way, 7 c 2 is going to be 21 So 21 into 16 you are absolutely correct barat Barat is correct. So barat can you just calculate 16 times 7 c 2 and uh type in your answer in the chat box No 37 is not correct ashi kosh So shan 37 is wrong Yes, it is 21 times 16 which is going to be 3 36 if i'm not wrong Right that is the correct answer. So 3 36 is the right answer guys Very good. I was not expecting this to come from your side, but well done Asutosh Sukirt Omkar Barat very good Okay, now see how it works very simple See when you say these two Should be simultaneously true. You are actually trying to say that x 1 plus x 2 should be 15 x 3 plus x 4 plus x 5 should be equal to 5 Okay, what I've done is I have made them Two separate events mutually exclusive from each others So I have to find simultaneously solve these two equations Now what is happening in this case? n is 15 R is 2 And of course when you say greater than equal to zero means you are saying blank groups are allowed blank groups are allowed Right So your formula is going to help you out here That is going to be n plus r minus 1 c r minus 1 Which is going to be 15 plus 2 minus 1 c 2 minus 1 Which is going to be 16 c 1 which is going to be 16 in the same way here n is 5 R is 3 And again blank groups are allowed Use the formula n plus r minus 1 c r minus 1 Which is 15 Plus 3 minus 1 c 3 minus 1 I'm sorry 5 Not 15 5 I'm so sorry. Yeah So it's going to be 7 c 2 And since this is ended Since this is ended You are going to multiply this so your answer is going to be 16 times 7 c 2 which is 16 times 21 which is going to be 3 36 Or 3 36 solutions you can call it. Yeah, Ashutosh Addition will not work because both of them have to be simultaneously be true if one is true Then it doesn't serve the purpose So remember fundamental principle of multiplication So both the tasks must be simultaneously being done Okay Guys, I will complicate the question with a bit more. Adwet you have any questions means multiplication adwet The logical way of thinking of and is this and this both have to be true So the number of ways would be multiplication of these two If I had said or Then I would have added it Is that clear adwet or these are slightly higher than mains But I'll be going to advanced level questions also in some time So let us move on to the next question again same type of question, but slight twist is there find the number of Find the number of non negative integral solutions integral solutions of 3x Plus y plus z equal to 24 Okay, so read the question very carefully number of non negative Integral solutions and this 3 is also present over here or a change Yes, it becomes n plus 2 minus 1 c 1 ashutosh should becomes n plus 1. So you're correct Guys think carefully and try to solve this No ashutosh 28 c 5 is not correct Take some time take some time Request all of you to give me the answer in Exact figure rather than giving in terms of c and all Give me an exact figure take a minute everybody try it out Because understanding of such concepts is very very important for you Sukirti you are absolutely correct answer is 117 Very very well done Yes, it is correct Awesome So just like in the class Sukirti is spoiling the game for everybody, right He's typing in the answers before others Don't look at the chat box guys focus on your notebook.  I'm waiting for you. Don't worry Yeah, uncle. Sorry 9 9 answer is 117 actually Trying with multinomial it may take a bit of time for you, but uh, you know, yeah You can do it by multinomial as well But there's another method without multinomial also that I'll be discussing in some time I'm just giving you one minute to try it out Keep posting your answers. It may be wrong. It's fine. No problem Giving a wrong answer is better than not trying at all Yes, Bharat, you're correct 117 is the right answer All right, so those who have uh, no clue how to solve this Let me just explain it to you how it works So whichever variable has the highest coefficient. I think this has the highest coefficient 3 what I'll do is I'll take that as k Okay And let's say k sum given number of course k cannot go beyond 8 Yes or no Because you have a restriction of 24 as your total sum. So your k cannot go beyond 8 So it is somewhere between 0 to 8 Correct So now what I'll do is I'll write the entire equation as 3k plus y plus z equal to 24 Okay So y plus z will become 24 minus 3k Right And of course your x y z all can be greater than equal to zero. So think as if Your n is this Your n is this 24 minus 3k Okay, and r is 2 So you're distributing these many identical objects between these two guys y and z So what does the formula say formula says it will be n plus r minus 1 cr minus 1, right There's a number of ways of doing it. So I can write this as 24 minus 3k minus 1 plus 2 c 2 minus 1 Which is going to be 25 minus 3k c 1 which is nothing but 25 minus minus 3k Okay Now you are trying to find out the number of solution for this. So basically you are trying to say that You're trying to say that If k is equal to zero your number of solution is going to be 25 Correct if k is going to be 1 your number of solution is going to be 25 minus 3 If k is going to be 2 your number of solution is going to be 25 minus 6 And so on till you reach k is equal to 8 Right for that your number of solution will be 25 into 3 into 8 Correct isn't like an arithmetic progression starting from 25 22 Then 19 all the way till you reach 1 Correct. So just add them, you know your sum of a ap Right, it's n by 2 2 a plus or you can say n by 2 first term and last term Right. So altogether nine terms are there. So 9 by 2 1 plus 25 would be your answer Which is nothing but 9 by 2 into 26 Which is 9 by 2 into sorry 9 into 13 which is going to be 117 Well, then guys Anybody who has used multinomial for this Just write i on the screen if you have used multinomial for this Good Ashutosh use the same thing. What about others? Did anybody use multinomial for this? Nobody All right, great You could have also used multinomial here guys in multinomial. I'll just give you a hint how it works So if you want to solve the same thing by using multinomial theorem, please remember that For the first distribution, you'll have to use x to the power 0 x to the power 3 x to the power 6 Like that till x to the power 24 Okay For y and z this is for 3x for y and z you can start from 1 up till 24 Like that And here you are searching for coefficient of x to the power 24 in this So you can try this out After the class, please try this method out and do confirm whether you have got the answer Only take third powers. Yes, Bharat. We are only taking third powers for the first one because it moves in a power of 3 You can either have 0 3 6 9 12 15 18 21 24 While the others this is the distribution for 3x This is the distribution for y. This is the distribution for z Right great. So please try this out guys and girls And do reach out to me if you get the answer with the multinomial, you should be getting it Moving on any question so far Guys, I'll give you a break at around 245 after this problem So if you are waiting to eat something or go out to have some water or tea, I'll give you a break at around 245 So let's move on to the next question in how many ways In how many ways an examiner can an examiner Assign 30 marks Assign 30 marks to eight questions to eight questions giving Not less than Not less than two marks To any question anyways can as examiner assigned 30 marks To eight questions giving not less than two marks to any question This should also be very very easy Just waiting for your answer No shushants. It's not 13c7 Yes, Andrew is correct. Sukirti is correct. Ashutosh is correct Bharat You're wrong. It's not 13c7 Yeah, I thought it was a very simple question. I don't see you know time being Taken in this question So first give two marks to every question, right? So you have lost 16 marks So give or assign two marks To each of the eight questions Which means you have lost 16 marks now So you're left with This many marks to be given to eight questions With no restriction now So r is 8 n is 30 minus 16 which is 14 So the answer is n plus r minus 1 c r minus 1 which is 14 plus 8 minus 1 c 8 minus 1 Which is going to be 21c7 Not less than two marks means Giving minimum of two marks to every question Yes, yes, shavan. He can give two marks for a question Never mind marath. It's okay. Next question is slightly different guys Please read this carefully. This is something different. I mean it is not based on whatever discussed so far A bag Has 23 balls in which Seven balls are identical In which seven balls are identical Find the number of ways Find the number of ways Of selecting 12 balls Of selecting 12 balls From the bag So shant, we are first we have first allotted two two marks to every question So there is no way that a zero marks can be given And then whatever marks is left off that is let's say 14 marks is left off There I can assign zero to any question because every question has already got two marks Oh my god marath Relax You can leave your answer in terms of c You can leave your answer In terms of c This is also a simple question, but it may take some time or let me give you this uh uh Let me give you the answer so that you can prove it. The answer is 18c 8 plus 18c 6 how Or show it No marath it doesn't work that way This is what is the answer 18c 8 plus 18c 6 That's absolutely correct. Sukirti is correct, but Sukirti I want you to simplify it. You are correct. Your expression is absolutely correct I want you to simplify it and bring it to 18c 6 plus 18c 8 Yes, you are correct Sukirti you are correct. Okay. This is just a basic understanding of your very very Basic core concepts of permutation combination. He has to select 12 balls Okay, and seven balls are identical Okay, so he may end up You know selecting the following cases number one Let's say identical and distinct combination. I am making it. Okay, so he can select No identical balls and all the 12 distinct balls And this can be done in 23c 12 ways I'm sorry 16c 12 ways Okay, because 23 balls they would be seven identical And 16 distinct second scenario would be One identical 11 distinct which is nothing but 16c 11 There's no there's there's just one number of ways choosing the identical ball So you can say one into this Third situation is two identical ten distinct which is again one into 16c 10 And can go on all the way till Till you have seven identical balls and five distinct Right, that is case number eight So the number of ways is doing is one into 16c five Why is this damn hard? You just have to understand in order to choose identical balls. There is only one way For example, if I want to choose a three identical from From seven identical There is just one way of doing it Because identical balls, there is no distinction you pick up any three balls from the seven that is going to be just one way of doing it right This 16c 11 16c 10 etc. These are only for the distinct balls When the balls are distinct Then only ncr is used So ncr is to be used when n is distinct ncr doesn't work for identical balls it doesn't work for identical objects So this is the basic that you need to know As we don't worry it comes with practice You know, even when I was a student I used to find it difficult But don't give up just keep practicing from basic books first. Just complete ncrt first do adi sharma Try this engage module You'll start getting confidence in it, but make your basics very very strong. You cannot mug up in this chapter There's no way you can mug up in maths at least Okay now The idea is how to Simplify this because most of you have got the answer which is nothing but your answer is 16 c 5 16 c 6 16 c 7 16 c 8 16 c 9 16 c 10 16 c 11 and 16 c 12. How do you simplify it? Can somebody tell me which identity will you use to simplify this? Pascal's identity try to recall Pascal's identity Recall Pascal's identity. What is Pascal's identity? Pascal's identity says ncr plus ncr minus 1 is n plus 1 cr use this Use this over here, right? Cello, I will help you to simplify this Let me just clear this off. I don't want extra things on the screen right now Right Yeah, tell me what is 16 c 5 plus 16 c 6 This is going to become 17 c 6 Right by using this Pascal's identity Okay In a similar way, this is going to become 17 c 8 This is going to become 17 c 10 And this is going to become 17 c 12 Right Let's see what I'm going to do here 17 c 6 can be written as 17 c 11 And 17 c 8 could be written as 17 c 9 Yes or no So what I'll do now What I will do now I will Club these two 18 c 11 and I will club these two I'm sorry One second one second just just one second Yeah, I'm going to club these two And write it as 18 c 12 And I'm going to club these two and write 18 c 10 So 18 c 12 is as good as 18 c 6 Which is what is the term over here? So I got this term And 18 c 10 can be simplified as 18 c 8 which is the term over here Right, so hence Hence shown Is that clear? Yes, I know most of the time is going to take in the simplification But see this is how it works many a time in the exam you have to match your options So you need to bring it exactly to the form in which the options are given to you All right, now you guys you can take a break for around 5 10 minutes Will resume exactly at Let's say 255 So let's have a break Okay We'll resume at okay fine is 255 three. It's okay. I mean somewhere between 255 to three will resume Yeah So please get yourself a bit of food or water And then we can extend to another One one and a half hour of session. We have a lot of things left off A way we'll try to have it somewhere between 434 45 Till wherever our requirement is because uh, we have some more parts to cover in fact Now more serious things will come up So we'll we'll see accordingly. Okay Guys, welcome back. So those who are ready in front of the screen Let us start Okay, so I've been getting requests to Complete the class by 415. Okay, so we'll end it at 415. No worries So And all the best for your exam tomorrow also Let us resume here and I'll be starting with You know A question in this in fact, it's a theory which I want to start with a question So that as we are solving this question We uh learn the theory as well Okay The question goes like this find the find the number of positive Integral solution Of the equation Of the in equation in fact x plus y plus z Less than equal to 30 Okay So read the question very very carefully. It is asking the number of positive right integral solution Of the in equation. So it is not an equation anymore. We have an inequality sign coming over here So I have to find the number of positive integral solution of this in equation x plus y plus z less than equal to 30 So those who are back welcome back We are starting this session now with a simple problem. In fact, not a simple one to look at Find the number of positive integral solution of the in equation x plus y plus z less than equal to 30 So basically this is a different type of problems that has been asked in j nowadays where you are not given an equation You are given an in equation Okay, how do I solve this question? That's correct. So Keith that is correct. All right. So whenever it comes to in equation guys Yes, you add a dummy variable. So in this case, we follow the approach of a dummy variable Okay Once again, so Keith did you take care of the fact that it should be positive? I think that mistake would have been done positive is what is the key over here positive integral solutions positive means x should be greater than equal to 1 y should also be greater than equal to 1 And z should also be greater than equal to 1 Okay All right. So let's take that into account now Now guys when it comes to inequality, what do we do is? We add a dummy variable over here. Let's say k that is what so Keith is saying add a k k is a dummy variable Okay, dummy variable means it was not there in my problem, but I introduced it Okay And k can be greater than equal to zero There's no restriction on k except for the fact that it should be greater than equal to zero Okay So whatever this quantity was less than equal to 30, right? So I added a k to make it equal to 30 Okay Now again, just like I told the restriction has to be met over here greater than equal to 1 For x y and z and just greater than equal to zero for k What I will do next is I'll replace my small x with capital x plus 1 small y with capital y plus 1 And small z with capital z plus 1 So this entire thing becomes capital x plus capital y plus capital z plus 3 plus k Is equal to 30 Which means capital x plus y plus z plus k is equal to 27 So here Now all the quantities involved are greater than equal to zero. So all x y z k here are greater than equal to zero So this is like saying blank groups allowed Yes or no So your n is going to be 27 r is going to be 4 And we are going to use our very old formula which we learned today 27 plus 4 minus 1 c 4 minus 1 which is going to be 30 c 3 Well done Ashutosh Bharath Sukheeth very good So please remember this concept of dummy variable, which is going to be very handy in solving inequality problems Okay Now a reverse situation of this. Let me take another situation a reverse situation of this Any question guys and girls, please free to type it in your chat box. I can read your questions. Don't worry about that reverse question is In how many ways in how many ways can We get a sum of at most At most 15 By throwing By throwing six distinct Dice Okay So read the question carefully. We need to get a sum Of at most 15 By throwing six distinct dice Think over it for one minute and if you've got the answer, please type the answer in terms of the exact Figures not in terms of ncr Okay This is a very good question where you will see that you will have to use multinomial theorem five Five thousand five. No, that is not correct. Ashutosh. Why it is not correct.  Okay. See how it works now Your every die can show you numbers from one all the way till six It cannot show you beyond six, right? No die can give you a number beyond six and know what die can give you a number lesser than one And there's six such dice Now what I have to do here is that What I have to do here is that I have to first before we write this I have to first Understand what I'm going to do here Let's say Let's say the sum which is obtained on die number one Let's say die one the sum of the number obtained is x one Die two the number obtained is x two I like that Till die six the number obtained is x six Okay Hold on hold on Bharath. Hold on. There are a lot many things to it So what is the question saying the question is saying that The sum of these numbers should be at most 15 means less than equal to 15. Are you getting this point? That means here you have to use dummy variable Here I have to use dummy variable Okay So if I use a dummy variable, let's say x seven Then I can say by using the dummy variable. I can write this as equal to 15 equal to 15 It is not 8 25 When I'm trying to solve this, let me just clear this off. I'll write it again once Here you have to take care of the fact that x one x two all the way till x six These numbers have to be between one and six right whereas x seven should be greater than equal to zero Because it's a dummy variable. It has to be greater than equal to zero So guys and girls I want your hundred percent attention on the screen now In order to solve this problem, we are going to find the coefficient of First x to the power 15 In x plus x square x cube x four x five x six whole to the power of six And one plus x plus x square all the way till infinity. Is that clear? So this case becomes a case of pure multinomial theorem So why why is this clear? Why we have written x plus x square x cube all the way till x six whole to the power of six It's because of the six die which you can Which can show only numbers from one to six And this is because of the dummy variable This is because of the dummy variable Yes, shawen. That's what I'm explaining the second bracket is because of the dummy variable which can take values greater than equal to zero Got it great. So in the next step, I'm going to just help you solve this This is very clear this step if it is not clear then there's no point solving it further So please please feel free to ask me any question at this stage Why from one to six y to the power six and why the other one is starting from one going all the way till infinity Please ask me Fine. So I think there's no question. Let me just solve it moving on to the next page So we are again, I'll write it down. We are again finding the coefficient of x to the power 15 in x x square Till x to the power six whole to the power six One plus x plus x square all the way till infinity Now I can make my life simple over here. I can use a geometric progression here Okay, where first term is x Common ratio is x number of terms is six By one minus x Whole to the power of Six and this is just one by one minus x So we are finding coefficient of x to the power 15 in this term Is this clear to all of you? Please keep it interactive if it is not, please Write it on the chat box All right. So now what I can do is I can clearly take out x to the past six out Okay one minus This to the power six And this will combine to give you one minus x to the power minus seven So coefficient of x to the power 15 in this Clear so far. I've just simplified it Okay Now shushant just for you If there is x to the power six outside waiting and we want x to the power 15 Can I say it is equivalent to saying It is equivalent to saying coefficient of x to the power nine in One minus x to the power six to the power of six One minus x to the power minus seven So I dropped x to the power six from both the sides So shant, where are you? Is it clear?  Is this clear shushant? Good. Okay Now guys all of you pay a lot of attention over here because things are going to become complicated here. Okay Now see this term which is one minus x to the power six if you expand it You are going to get one minus Six x to the power six. Hope you know a bit of binomial expansion Okay The next term is going to be plus Six c2 which is going to be 15 x to the power 12 There would be more terms here, but I would not write further even I should not write this term also because it has already exceeded nine Uh, no, no, no.  Barats.  Okay It's a normal die that we normally use to play our games of chess or Whatever you want to call it ludo Guys, is this clear? So when I'm expanding this term, which is only this term I'm not going to write any further than these two terms I should not even write this term also because it goes to a power of 12 after that So it is not going to give me x to the power nine anyways Right, so I need only the first two terms now guys this term which is One minus x to the power minus seven This term is a infinitely long term This is an infinitely long series okay, so Yes, I'm now going to use the formula that we derived there How I'll tell you in some time So basically we are looking for coefficient of x to the power nine in this Right, let me write it in a much simpler way We are looking for coefficient of x to the power You know nine in this Okay Now guys, let us use our common sense over here If I want to generate an x to the power nine Okay x to the power nine can be generated When this one over here Let me choose a different color right now We choose a blue When this one over here Meets x to the power nine from here So we need x to the power nine from this term Okay And x to the power six meets x to the power three from this term So just cater to the requirement of the problem. So I need or we need x to the power nine from In fact x to the power nine And x to the power three from One minus x to the power minus seven. So I'll just Pick out these two terms only. Yes, Bharati. You're correct. We need x to the power three and x to the power nine in that term So I would not bother to expand it completely I would save my time and energy. I'm not going to expand it completely I'm just going to hand pick these two terms From this term Over here that is one minus x to the power minus seven Let's see. How do we do it? And as Sukrit already pointed out, we are going to use our readymade formula Okay, what was the formula? I'm sorry. What was the formula that we learned? We learned that coefficient of x to the power n in one minus x to the power minus r What is this formula? Please type it out in your screen n plus r minus one c this right Please type in the formula coefficient of x to the power n in a one minus x to the power minus r I've already written it out. Okay Or you can say n plus r minus one c n correct Okay, now let us find out Again, let me write the problem statement coefficient of x to the power nine in one minus six x to the power six one minus x to the power minus seven So now I'm focusing on getting x to the power three from here. What is the coefficient of x to the power three from here? So n is three r is seven minus one c three correct Correct, which is going to be nine c three. Similarly, I would find out coefficient of x to the power nine from this What is going to be that? Again, n is nine r is seven minus one c nine Which is going to be Which is going to be 15 c nine Correct. See guys, what I'm doing I have to find out coefficient of x to the power nine and x to the power three, right? That is what I'm going to do now So I'm trying to find out that only from this So x to the power three I have got x to the power nine I have got So what I'm going to do is I need coefficient of x to the power nine in one minus six x to the power six There would be many terms here boys and girls But I need these two terms only These two terms are only required Rest of the terms are useless for me. So I'm not going to even bother writing them Correct So my answer is going to be when this one multiplies with this I get x to the power nine correct When this multiplies with this I get x to the power nine So coefficient of x to the power nine would be 15 c nine into one minus six this six Into nine c three which is this are you getting it? 15 c nine is five thousand five this figure seems familiar to me somebody. I think ashutosh I think ashutosh typed this particular thing on the screen. Okay And six into nine c three is going to be five hundred and four you can just calculate it Okay, nine c three is nine into eight into seven by six which is going to be 84 84 into six is going to be five not four So your answer is going to be five thousand five minus five hundred and four Which is four five zero one This is going to be your answer It's a very very good question. We learned a lot of things from this question It's a very very good question three star many things we learned how to formulate A multinomial expression how to use dummy variables how to use Or how to expand binomial terms having negative integral powers How to find coefficient of any power of x in that infinite series expansion And finally how to you know get simplified these things See ashutosh if you look for alternate ways those alternate ways would end up giving you so many cases So that is not going to help our cause You need to be you know very comfortable with this approach rather than finding alternatives to it Okay. Yeah, shavan. I can give you this same type of question Uh, but this time Okay, I'll I'll just tweak the question a bit So this question was getting a sum of at most 15, right? Now let's say I asked you a question In how many ways Can you get a sum of greater than 15 greater than 15 By throwing six distinct normal I have to introduce this word normal because of barat normal dice Okay, how many ways Can you get a sum greater than 15? By throwing six distinct normal dice Try this out guys after this we have we have to uh Do uh derangements as well So I wanted to finish it off before 4 15 so that you get enough time to study for your exams tomorrow Yes, ashutosh you are correct.  This should be greater than 15. How do you deal with this? No ashutosh The answer is much much more than that. Anybody can give me a clue How do you deal with a situation greater than 15 for less than equal to you can use a dummy variable For equal to also, you know, how it works What to do when it is greater than 15? minus k Okay, so here the hint is guys hint is It's total number of solution minus The number of solutions where you get This as less than equal to 15 because we know total We know less than equal to okay There's the most simpler way to simplest way to solve this problem. So total number of ways is 6 to the power of 6 isn't it Because a dice can show each dice can show you six expressions six results And this we already found out was five four zero one Okay So if you use your uh calculations, this comes out to be This which is about four two one five five Right Bharat. Correct. It is six to the power six minus the previous answer. That is what precisely I did over here And this is our final answer So guys, just remember this hint. This is a very very important hint When you're solving for such in equations where you are asked to solve greater than that And of course there would be an upper limit to the answer So it's total minus less than equal to that will automatically give you this Yes, four to one five five is the answer So one last question I'll take on this and then we'll move on to derangements What he said initially was not clear to me but later on he said six to the power six minus the previous answer So that is what I also said Six to the power six is the total cases So every die has six outcomes. So there are six such dice So we'll have six into six into six into six six times. So six to the power six outcomes minus the situation where the sum is less than equal to 15 We know the number of uh ways in which you can get a sum less than equal to 15 is four five zero one So the difference of it will give you the answer Okay next question is the number of positive integral solution x y z such that x y z is equal to 24 Yes, we're precisely we're seeing the same thing. Yes. Is the question clear? Okay So we have to find the number of positive Integral solutions x y z such that the product of x y z is 24 Please try this out for one minute 30 is correct Away your correct 30 is the right answer very good. Yes as though it is correct 30 is right 5c 2 into 3 is absolutely correct again. It's very simple you first have to Prime factorize 24, which is 2 to the power 3 into 3 to the power 1 Now think there are three boxes x y z And you have Three objects of type 2 So you have three objects of type and one object of type three So think as if two is a type of object like let's say two is a banana Okay, two is a banana three is an apple So you have three apples and you have one banana and you want to put them In three boxes or you want to distribute it between three people Okay, it becomes as good as saying that you want to distribute uh three people x y z Three bananas and again x y z you want to distribute one apple And these two are ended Are you getting this point? so guys Okay, so Keith. Hope you are able to connect to us through your phone if your computer crashed Okay, so guys this is called making an analogous situation So learn how to make analogous situation This is something which you will come with a lot of practice Okay, now this is very simple here n is three r is also three And blank groups are allowed of course So it will be n plus r minus 1 c r minus 1 which is going to be 3 plus 3 minus 1 c 2 which is 5 c 2 And this is n is 3 I'm sorry n is 1 n is 1 r is 3 So it's going to be n plus r minus 1 c r minus 1 Which is going to be 4 minus 1 c 3 minus 1 so this is going to become 3 c 2 this is going to become 5 c 2 So your final answer is going to be 5 c 2 into 3 c 2 Which is 10 into 3 which is 30 ways or 30 solutions Is that clear So guys the key is this The key is this your situations may vary But if your core principle is intact you can make analogous situations to solve this problem Is that okay Is this clear to everybody how it works Yes, it is the Star and bus algorithm that is going to work here Great, so now we are going to move on Now we are going to move on to the last segment of this chapter which is Called derangements And of course, we'll have a lot of problems after that We have not yet done problems based on counting the number of squares and rectangles So i'll be doing that also after i finish this off and if i get time That is laddus and sticks barath same thing Okay, so what is derangement First let us understand what is derangement Derangement is basically defined as any change in the order of the things in a group So let's say when n things When n things Are to be placed Are to be placed at n specific n specific places But none of them But none of them Is placed On its specific on its specified position On its specified position Then we say n things are deranged n things are Deranged Right To understand a very simple example of derangement is let's say There were four letters letter one Letter two Letter three And letter four Which are to be placed in four envelopes So these envelopes are basically addressed as per these letters Okay, so the specified position for The specified position for this was this specified position for this was this for this was this and for this was this But let's say no letter goes No letter goes To its Designated envelope Designated envelope Then it means we are deranging these four letters Or you can say four objects Okay, you can derange any two of them also Right So now we are going to talk about Bharat, there's a lot of things involved here In fact, we are going to derive the number of ways how we derange n things among themselves So is this definition clear to everybody has everybody understood what is the meaning of deranging r or deranging n objects Please write clr on your chat box if it is clear So none of them is placed in their specified position Shavan Andrews clear good. So now I'm going to derive the expression We have we are going to derive Let's say formula for for derangement of n objects Now listen to this It is again through that inclusion exclusion principle which we had discussed Inclusion exclusion discussed While we were doing the distribution of n different things in our different groups such that blank groups are Not allowed Okay, so I'm going to revisit this principle once again for deriving the formula for derangement Yeah, Ashutosh. What I'm saying is I'm going to use the inclusion exclusion principle once again that I used in the class while I was doing Distribution of n different objects in our different groups When blank groups are not allowed Okay Now how it works I'll take an example to explain this exclusion inclusion principle Let's say There are n letters like this There are n letters letters means Chitty not alphabets, okay And there are n envelopes Okay, n envelopes Okay Now listen to this very very carefully Let's say AI be the event Let AI be the event that The i-th letter letter Goes Inside or goes in the i-th envelope That means this i-th letter goes in its correct envelope the one which was You know designated for that it goes in the that correct envelope only Okay Now you tell me if i-th letter goes in the correct envelope How many letters remain n minus 1 And in how many ways can I shuffle these letters among themselves? You would say the number of ways in which you can shuffle these letters among themselves would be n minus 1 factorial Now guys, I'm not saying here that Any other letter is not going into its right envelope Any other letter can go in its right envelope? It cannot go also it may or may not go But it is for sure that the i-th letter is going in its correct envelope i-th letter is going in the i-th envelope other letters Other letters May or may not go in their respective envelopes are you getting this point? Okay So I can say that the number of ways in which i-th letter goes in its correct envelope would be n minus 1 factorial Because you are just trying to shuffle the remaining n minus 1 letters among themselves Is this clear in a similar way? Okay Let a i intersection a j be the event be the event that The i-th and the j-th Letters Go in their respective In their respective envelopes so remaining Remaining n minus 2 letters can be shuffled among themselves in n minus 2 factorial ways Is that clear? So similarly, I can take this and say n a i intersection a j intersection ak could be done in n minus 3 factorial ways and so on Right and so on so far. Is it clear? I'll repeat the process once again let Let a i be the event that the i-th letter goes in the i-th envelope. That means the other n minus 1 can be Shuffled among themselves in n minus 1 factorial ways Then again I took another case let a i intersection a j be the event that both these letters go in their respective envelopes So remaining n minus 2 could be shuffled in n minus 2 factorial ways And so on so if three letters are going in their correct envelopes Then remaining n minus 3 could be shuffled in n minus 3 factorial ways and so on Is this clear so far? Please type clr if it is clear so far Essentially, I'm repeating the same thing which I did while doing the inclusion exclusion principle for the case in the class. Others please type clr if you feel the thing is clear to you Okay, great Moving on Now what is the total number of ways? That is if I say what is the universal set? You would say n factorial right because n letters can be inter arranged in n factorial ways correct Now guys, what am I looking for? I'm looking for that case Where the first letter doesn't go in the first envelope And the second letter doesn't go in the second envelope And the third doesn't go in the third envelope and so on till all the n letters right Which as per the d morgan's law is nothing but n u minus n a1 union a2 union a3 all the way till a n correct by d morgan's law Which is nothing but Which is nothing but n union Sorry number of elements in the universal set minus summation n ai minus summation n ai intersection aj Plus summation n ai intersection aj intersection ak minus sorry Minus and so on so this is something which we already know from our set theory chapter Now what is If n ai is n minus 1 factorial What is summation n ai? In how many ways can you pick up that ith letter you would say it is nc 1 times you can pick up And then n minus 1 factorial it is the same expression. Yes In a similar way if n ai intersection aj Is n minus 2 factorial What is summation n ai intersection aj? So how many ways can you pick up aj out of n you'll say nc 2 ways And then put them in the right envelope And the remaining n minus 2 you can shuffle among themselves in n minus 2 factorial So like that it keeps on going So let me put it back in the expression number one So replacing it in expression number one or substituting it in one n universal set is going to be n factorial minus nc 1 n minus 1 factorial minus nc 2 n minus 2 factorial Plus nc 3 n minus 3 factorial minus nc 4 n minus 4 factorial All the way till the end Okay, the last term would be In fact, let me write it down Or let me open this up and then I'll let it down Yeah What is nc 1 into n minus 1 factorial it is n factorial by n minus 1 factorial into n minus 1 factorial Okay, so these two terms will get cancelled off So this term over here This term over here is just n factorial Okay, let me check this term now This term is n factorial n minus 2 factorial 2 factorial into n minus 2 factorial So these two terms will get cancelled So this term over here Is n factorial by 2 factorial Similarly, this term here is n factorial by 3 factorial This term is n factorial by 4 factorial and so on Okay Now let me expand this Let me just clear off this part When I expand this I'll get n factorial Minus n factorial Plus n factorial by 2 factorial Minus n factorial by 3 factorial Plus n factorial by 4 factorial minus n factorial by 5 factorial and so on Correct me if I'm wrong This will go to minus 1 to the power n n factorial by n factorial Correct If I take an n factorial common It becomes 1 minus in fact just to maintain consistency. I would write 1 factorial over here So it becomes 1 minus 1 by 1 factorial Plus 1 by 2 factorial Minus 1 by 3 factorial Minus 1 by 4 factorial And all the way Once again once again once again one sign mistake Plus yeah All the way till minus 1 to the power n Into 1 by n factorial Okay Okay, so let me write it out in a clear space so that all of you can read it properly So the number of ways in which you can derange n objects is given by n factorial 1 minus 1 by 1 factorial Plus 1 by 2 factorial Minus 1 by 3 factorial Plus 1 by 4 factorial And so on till you reach Minus 1 to the power n 1 by n factorial It's a very very important formula very important formula. Please keep this in your mind Okay, please remember this formula So let us go back to the example of the four letters and four envelopes So how many ways can I derange four letters? So that no letter can go to their respective designated envelope So your answer would be put n s 4 So 4 factorial 1 minus 1 by 1 factorial 1 by 2 factorial minus 1 by 3 factorial Plus 1 by 4 factorial Let's calculate it These two terms anyways would get cancelled So it's 24 half minus 1 by 6 Plus 1 by 24 Which is going to be 12 minus 4 plus 1 which is going to be 9 ways Is it clear? so guys This is the formula for derangement of n objects taken all at a time So before we start taking more problems, there are certain properties that we need to Keep in our mind that will help us save our time First property is you should know these values For example D1 is 0 Now how simple If there is one letter and one envelope You cannot derange them Right That letter has to go in its respective envelope. So D1 will be 0 And you can anyways check from the formula also D1 would be 0 D2 would be 1 Okay D3 is going to be 2 D4 is 9 just now we discussed that D4 is 9 There's some values which you should remember D5 Is going to be 44 D6 is going to be 265 Just remember them because it will save your time They're not very difficult to find out. You just have to use this formula over here start putting your d as n as 3 4 5 6 you'll start getting these figures If you remember this figure it really really saves a lot of time Okay Next property is Next property is property number two dn plus 1 Is n times dn plus n times this So for example, let's say I know d2 I know d3 I know d4 Sorry, I know d2 d3. I want to know d4 Okay, very simple put n as 3 over here. So d3 will be 3 times Sorry d4 will be Put n as 3 d4 would be n times which is 3 times d3 plus d2 Just now we learned the value of d3 and d2. What was d3 and d2 guys? What was d3 and d2? D3 was 9 And d2 was 2 right Is that clear? Yes or no So d3 was sorry d3 was 2 Yeah, d3 was 2 and d2 was 1 So as you can see we got d3 as 3 into 3 which is d4 as 3 into 3 which is 9 Which is what we dealt in the previous slide If you look at the previous slide d4 was 9 Right here d4 was 9 Okay So please remember this a useful property. I mean may not be very useful in solving problems, but yes if you know it You can save a time in some places third property third property Is dn plus 1 is n plus 1 times dn By the way, this doesn't work when n is n should not be equal to 1 in this case Okay Yeah dn plus 1 is given as n plus 1 dn Plus minus 1 to the power n plus 1 You will prove this for homework I want to see the proof for this and if you're able to get it, please uh post that on the group Okay, so please post the solution for the derivation of this third property that dn plus 1 is n plus 1 dn Plus minus 1 to the power n plus 1 Okay Guys before we wrap up this session. I like to take a question on this Okay, let me take few questions on this first question is in how many ways In how many ways Can six couples Okay six couples Play a mixed singles mixed singles such that Such that no couple play against each other No couple Play against each other Hope you all understand the meaning of mixed singles means One man and one woman have to play this Let's say badminton baddie Okay So man and woman should not be husband wives Ashutosh you have to practice So basically this is a very simple question and the answer is already there in front of you most of you have already replied It's actually d6 Because if you don't want any couple to play against each other, it is like, you know, you treat this as letter and this as envelopes No letter should go into their respective envelope. That's it So the answer is going to be 265 Yes, it is d6. Correct. Sushant. It is not six to the power five It is d6 So The question may be given to in any format. It is up to you to realize that oh, I can use my concept of derangement over here Okay Next question, which is I would say the last question for the day and then you guys can go and study chemistry Okay question is a person writes Letter Or letters to six friends to six friends Okay in how many ways In how many ways He plays these Six letters in six envelopes such that letters are in wrong envelope Is the question clear? So six letters six envelopes At least two letters should be in the wrong envelope. Please solve this and type in your answer on the chat box Please type it in Figures don't give me expressions in terms of factorials When you're saying d6 means you're saying All the six are going in the wrong envelope. How is that possible? Oh, you're saying 63d6 Please check your answer again. You're on the right track though. No shawan. That's not correct Sukith no that is Oh, you're very close Sukith. Your answer is just off by one Let's check some mistake has happened answer is 719 Shushant, uh, your Intentions are correct. Your approach is correct, but the answer is not correct Yes, as soon as the answer is 719. Yeah. Yeah, correct 719 is the right answer guys Let's solve this So you guys very simple you want your at least two letters to go in the wrong envelope, right? So you can have following cases Okay, your cases can be let's say, uh Right envelope and wrong envelope. You can have Uh, four in right two in wrong three in right three in wrong Two in right four in wrong Right We can have we can have one in right five in wrong zero in right all in wrong Okay, that's how we take care of at least two letters going in the wrong envelope So we have five cases all together correct Now see how it works Let me go to on the next page. So let's say You want Four letters to go in the right envelope and two to go in the wrong envelope. So first of all choose The four which you want to go In the right envelope or in their designated envelopes Okay, how many ways can I do that? I can do that in four. Sorry I can do that in Six c four ways Two to go in there wrong envelope. So the remaining two have to follow d two So the number of ways in which I can do this case is going to be Is going to be six c four into d two Are you getting this point? In a similar way if you want Three in right three in wrong then it will be six c three into d three Now Some may somebody may think why it is just six c four. Why not six c four into four factorial? See guys, there's only one way to put those letters in the right envelope Once you select those four. There's just one way in which you can put those four letters in the right envelope. So just six c four It is not going to be six c four into four factorial And this d two means the remaining two are deranged among themselves In a similar way If you want two to go in the right four to go in the wrong. It will be six c two into d four Then one to go in the right five to go in the wrong. It will be six c one into d five And then zero to go in the right all of them to go in the wrong will be d six Correct. So let's calculate this six c four if I'm not wrong it is 15 And d two value is one six c three is 20 d three value is two 62 is again 15 d four value is nine Six c one is six d five value is 44 and d six value is 265 So answer is 15 plus 40 This is going to be 135 This is going to be 264. This is going to be 265 at them all Answer is going to be seven one nine ways Is this god level clear type glc if it is god level clear Good guys. So we are going to end up this session right now In fact, there are some more types of problems which I wanted to do with you like counting the number of squares in a chess chessboard number of rectangles uh, which Yeah, shira, what happened? I'm using macbook here for that. Okay So shira, uh, if you have any concerns, please please feel free to ping me on the group as well Okay, no problem. No problem Yeah, so kids, I know the answer is two one two nine six rectangles But when things become different it'll be taking different different shapes anyways, I'll Call it a day now over and out from centum academy here for the youtube live session Thank you all for joining in And I wish you all the best for tomorrow's chemistry exam. Do well guys. Okay make centum proud So over and out from this side Thank you very much. Bye. Bye. See you


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The Chartered Institute of Bankers of Nigeria, CIPN, has declared a net operating surplus of 1.37 billion Nair for 2023, marking the significant growth from the 837.94 million Nair recorded in the previous year. This was announced by the outcome, President of the CIPN, Dr Ken Okbalar, during the Institute annual general meeting in Lagos, plus the correspondent, Love Ikukuo Yudokun, was there and now reports. In the 2023 annual general meeting, MAGM, an election of new officers who will stay at the face of the Institute for the next two years. The outgoing President, Dr Ken Okbalar, gave an open remark which was preceded by a minute of silence in honour of their departed colleagues. Departees, colleagues, operate and mention them, Mark Osam, Professor Mali, is the manor before the demonstration. Presenting his report, Dr Ken Okbalar said the Institute hit the billion Nair Mark, achieving a net operating surplus of 1.37 billion Nair in 2023, which represents a 63.60% growth. The net growth of 7.8 billion in 2023, 9.1 billion in 2023. The current income ratio for the year 80 December 2020, still at 50%, down from 59.4%, in the current 1.8 billion in 2023. The President expressed confident that the CIBN will continue to work stronger financially despite the economic downturns. After the presentation of Fox and Figures, the external auditors affirmed that the financial position on the CIBN was in agreement with a bulk of organs. And it was commendation all the way for the outgoing President, Dr Okbalar and his team for piloting the affairs of the Institute in an unprecedented growth. I was surprised to find that this went shift in reduction in the issue. So I think management has a greater efficiency, and I also know that the Council has done very well. So this is commendable. I have called management to continue to do those things that they've done, but have led to this reduced cost reduction. So what does the report provide again? As the past President or if it's a reality to your heart, you will see that the strategy trust has been followed, and we are growing in use and balance. And I hope that the incoming President and his team will observe the use and look at it and follow through.  All the suppliers on ORAY-1G were elected as a new President of the Institute, along with other elected offices for the 2024-2026. Say no. For the post of the President, we have Professor ORAY-1, two powers, OLAF-80, Ph.D., FCI-D, the $600,000 grant. In his acceptance speech, ORAY-1G described the outgoing President as a great leader who had positioned the Institute in a path for good and good means. He said at the same path for us, as we are preparing for this transition, that a very good man, that a very good man, will remain in trust for a very long time. Because, you came in, you know, without people's trust, you wouldn't have been able to achieve, you wouldn't have achieved the support of this petition. We give thanks from the path of the Institute and what we are witnessing today. Once again, I want you to correct your dedication to duty. He thanked the leadership and members of the Institutes and promised to build open legacies of his predecessor, Love Ikuku Unidoku. Plus, TV News. Don't forget to hit the notification button so you get notified about fresh news updates.