 chapter binomial theorem binomial theorem very easy chapter very very scoring chapter right so I think we should be able to cover this in two classes I'll just give you an overview of this chapter so what are we going to study under this chapter for your competitive level exam point of view in school it is not there it has been removed from your school exams right so we'll only be studying it from the competitive level exam point of view so the first thing that we are going to talk about is the derivation of binomial theorem so this is a theorem how does it work okay so what is the derivation of it okay how come how come we had one class on this in school but it was not there in your CBC syllabus that's what Raja Ji Nagar students told me ma'am said it is important okay then she is definitely a very concerned and sensitive teacher because she knows that it is useful unlike many teachers who are not teaching at all guys one thing I would like to bring to your notice board has never said dad you should not teach it right board has said you should not test it okay and she is very correct to say that it is very important okay even when I you know last batch your senior batch 12 batch we never omitted anything from our syllabus even though CBC omitted a lot of things and see it paid off now the board's JEMain syllabus says nothing is omitted okay but while many coaching institutes they omitted the syllabus and they taught so that their effort is less we could have also done it happily but in the long run it is going to harm you a lot my second name is you clearing an exam my first name is you understanding the concept enjoying it and knowing it very well exam is a by-product if you if you are studying from a firm exam point of view your knowledge will be very short-lived it will be very ephemeral okay yeah it's not I'm not saying all this thing to show how great is Centrum Academy but yes we definitely believe there is some non-negotiables that understanding is the utmost priority for us all right so derivation of the binomial theorem the next concept is the concept of the general term okay middle term terms they can be middle more than one middle term so middle term or terms okay then we'll talk about numerically greatest numerically greatest binomial coefficient term and coefficient they sound very similar but they are not we'll we'll talk about it when the right time comes okay next is next is the application of binomial theorem application of binomial theorem sometimes some cases divisibility problem remainder problem finding the last digit second last digit third like that etc of a very heavy number so those applications are there with us for us in this chapter and finally we are going to talk about not finally second last you are going to talk about binomial coefficient series binomial coefficient series okay and lastly a bit of multinomial theorem what again so we did this in pnc yes you definitely did it in pnc but we have to talk about it in more detail binomial theorem and finally you have a bit of idea about i mean 15 or 20 minutes idea on binomial theorem for for for non-whole number index okay so these are the seven things that we are going to study in this chapter even though it looks like seven things but i'm sure we'll be able to cover at least the first third in today's class on you okay and four five six seven can be done in the next class okay so are we ready to start ourselves with binomial theorem not a new concept you have been using this in our we have used this in our limits chapter when you were doing one of the standard limits called the standard algebraic limits so not a new concept so let's start first of all binomial theorem it was discovered or invented you should say by by newton one second aakash these are chapters topic names something which are not important like you can copy it after the session has comments sorry chat session is over okay front of which teacher is there for you in your school hsr all right so binomial theorem before we derive this binomial theorem there is certain observation let's let's do a certain observation okay important observation so i'll be introducing you to such terms where n is a whole number now n can be a negative integer or a fraction also that is something which will be taking in the end of the chapter remember the last part of the chapter was binomial theorem for non-whole number index so right now we'll talk about the whole number index which is basically 90 percent of your uh you know concepts in your binomial theorem last 10 percent 5 percent concept will be based on when the power is not a whole number now why it is called a binomial theorem because it is applied to a binomial expression like this right now why normally it was called binomial because we consider that these two terms are dissimilar so they're two dissimilar terms because they are two dissimilar terms we use the word binomial okay but that is like you know very very uh you can say rudimentary form of binomial theorem and it was you know uh invented by uh Newton we actually can apply it to similar terms also right we can also do two plus three whole square by use of binomial theorem doesn't make any difference to uh to the approach okay so don't be like too much into like okay they have to be dissimilar yes when it was invented when it was on the rudimentary phase raw form people use it for uh dissimilar terms a and x are dissimilar term now again my aim is not to tell you that a is a constant and x is a variable no think as if a is a first term x is a second term and they're dissimilar from each other kind of right but don't take it very you know in uh uh exact form don't think like they have to be dissimilar terms they can also be applied to similar looking terms as i already told you a certain observations i would like you to have let me ask this as a question what is a plus x to the power zero let's start with the lowest whole number which is zero what is a plus x to the power zero one okay everybody's saying one what if i write one as one a to the power zero x to the power zero any problem with that we'll say no sir this is one only no problem okay good what is a plus x to the power one guys just have the good observation because i'm writing it out for you to have a good observation on this so it's a a plus x correct a plus x absolutely guy three can i write this as one a to the power one x to the power zero and again one a to the power zero x to the power one right banana is trying to understand what is going on yes absolutely banana okay are you i should write this in yellow it will look beautiful no art sir for beauty and all you raised it let's have consistently white and yellow white and yellow okay next is a plus x whole square what is it you'll say i mean the formula that comes in your mind is a square two ax plus x square correct what if i write it like this one a to x not two a one x one and one a not x two okay okay so don't worry about these numbers that you see worry about this a and x combination how it is like written worry about that look a pattern in that numbers i will talk about it in some time okay next a plus x whole cube these all formulas you have already done in your junior classes so you say a cube three a square x three ax square and x cube right what if i write it like this one a cube x not three a square x to the power one three a to the power one x square and one a zero x cube correct now when you look at these terms only the ones which i'm showing from the right side with a bracketed under bracket okay what pattern do you see there how are they listed down any pattern we are able to observe here no no they are not falling past this triangle past this triangle is followed by these numbers which i'll talk about in some time guy three huh these these numbers a and x they are basically sharing that power or sharing this n among each other or between each other yes or no yes absolutely number so if you see zero is shared by these two one is shared by these two so one zero zero one okay so it starts with the entire power n being taken by a and then what a does it shares one power at a time with x sharing is caring right so a starts with let's say it starts with the whole power x starts with zero then what happens a gives one power to x and again one more power so like that a starts giving one one power to x and slowly it becomes zero and x takes the entire power okay this is a pattern i would like you to observe over here as of now right now i know most of you do not know the formula of a plus x to the power four okay but if you look observe this pattern at least we can write down these expressions so let's say there is some number i don't know which number but i will have a to the power four x to the power zero again some number a to the power three x to the power one then you have a to the power two x to the power two then you have a to the power one x to the power three and then you have some number a to the power zero x to the power four okay so this observation actually helps us to write these terms at least okay now what about these numbers which i have not mentioned what about these numbers now these numbers come from an observation made by a french mathematician blaze Pascal and he realized that it comes from a pattern like this so this one that you see this one that you see here okay write it like this this one and this one you write it like this okay one two one write it like this one three three one you write it like this and he realized that when he was listing down these numbers okay the first number in the last number is always a one and the number in between comes from the addition of the top two if you add these two you'll get a two similarly if you add these two you get a three you add these two you get a three again okay so with this pattern in his mind he was able to construct the further rules so the next so that will come is one one and here i'll get a four this will be six this will be four again okay and these numbers will fortunately become the coefficients of this question mark or they will replace these question marks which i have put over here so it would become one one four six four one and this triangle that blaze Pascal came up with was later on honored after his name and it's called the Pascal's triangle the Pascal's triangle okay not a new concept to you we have already seen it when we were doing our permutation combination chapter i gave you Pascal's identity i gave you hockey stick identity they were all based on this triangle itself so thanks to this observation of pattern and thanks to this Pascal's triangle that now we'll be able to write down a plus x to the power any power we want of course not a very heavy power right now because you have to make triangles to that extent so i would request all of you take a clue from this and just give me what is a plus uh seven sorry a plus uh six a plus x to the power of six everybody please write this out 30 seconds 35 seconds 40 seconds you might take yeah i'll talk about binomial coefficient coefficient everything he wants you don't worry and how they are similar and how they're different at the same time write down but it doesn't take a lot of time no need to type just say done no need to type please okay punnab is done good punnab just hold your answer whatever is also done done done done okay good yeah so what i'm going to do is first i will write down those ax combination something a6 x0 something a5 x1 something a4 x2 something a3 x3 something let me write here something a2 x4 something a1 x5 and something a0 x6 okay so six came up as a power on a first with x power being zero and slowly slowly a started losing power one by one and x started gaining power one by one now to know these bracketed terms you have to take the help of pascal's triangle so pascal's triangle this will be one five ten ten five one one six fifteen twenty fifteen six one okay so this will become one six fifteen twenty fifteen six one right so you don't have to literally you know do a plus x to the power six by you know multiplying it to itself six times okay so thanks to this theorem thanks to this observation of pattern and this pascal's triangle that we are able to figure this out so later on newton he basically formalized it in the form of a theorem so what theorem he brought up and let us link the theorem that he found out with this particular approach so i'm going to the next page permission is there or anybody is copying i'm sure most of you would have written this okay can i go to the next page okay so what is the theorem now so newton what he did a plus x to the power n when n is in a whole number it is what it is basically a plus x into a plus x into a plus x like that it'll go n times isn't it okay now read addition as or and read multiplication as and like the way we used to do it in permutation combination so see when you are trying to multiply it okay you're going to get several terms okay yes or no so when you're going to expand it you're going to get several terms right now each of these term that you see let's say this is your term one this is your term two this is your term three and i don't know how many of you know how many terms will come okay so let's say this term number i do not know okay does anybody know what till what terms how many terms will i get from this anybody has an idea n n plus one n plus two okay many people are saying end okay there's one thing let's keep p question mark okay some of you have studied it they know it's n plus one terms okay should i write it down okay let me write it down okay n plus one terms will come okay you could also observe the pattern which i gave you just now you can just see how is your exponent and number of terms related to each other see now what i'm trying to do is everybody please pay attention here any term that you're formulating whether it is t1 t2 t3 till tn plus one every term is basically made up of either a or x coming from the this factor and either your x coming from this factor and either your x coming from this factor so read this like this as if this is an army to be formed by picking up people from these households so these are households let's say house one house two house three correct till house number n now let's say an army has to be formed and this army will contain either a or x not both that means every household has to contribute one member only one member whether a or x that is our call and every household must contribute one right so think i normally give this example in every binomial session class so think that there is a country which has got n households okay and every household has a man and a woman so a is a man x is a woman okay and the country all of a sudden has to go for a war okay so war has been called by the head of the country so now as a responsibility of every citizen or every household every household every household has to contribute one person okay not both because if both the people are going for the this thing who will take care of the house okay so either a or x will participate from h1 either a or x will participate from h2 either a and h will participate from h3 and so on right and every household must contribute a particular person right now let us say every household gives a then the army will be formed of all a's so that will be a to the power n right that is the first term which you which we used to write okay you may also write it in a better way a to the power n and no x so x to the power 0 if you want to okay let's say there's another way to make the army so this is the first way to make the army another way to make the army would be take a from n minus one households and one x from the remaining household what did i say take a from n minus one households let me write it in white only to maintain the pattern and take one x from the remaining household but now this one household who is contributing x that itself can be chosen in nc one way so many people ask them why are you choosing that house which is contributing x let's choose the house which is contributing a so that can be chosen in ncn minus one way both of them are same as we have learned in our permutation combination chapter okay so it can happen that x come x can come from household number one and rest households will give you a right or it can happen that second household gives you x right so the female member of the second household says i will participate in the war and the other household h1 s3 etc they give a like that now in the same way just keep thinking how many ways can you make an army where two of the you know family contributes to x and the rest contributes to a so that will be nc to a to the power n minus two x square right if this tent continues the last will be ncn a to the power zero x to the power n means all women of the family have decided to go for a war okay and i'm sure they will win it yes or no in order to make it consistent we let's have nc zero right so this is what newton figured out right so let me write it like this okay so these terms that we used to see in the pascals triangle were actually nothing but they were the ncr terms let me show you how this one i mean just i'll write on few of the okay this someone that you see it is actually zero c0 this is one c0 this is one c1 this is two c0 this is two c1 this is two c2 this is three c0 this is three c1 this is three c2 this is three c3 okay i think i already told you this in your combination chapter but just to reiterate the same thing okay so what newton figured out was a theorem actually so he said that a plus x so what's the theorem so the theorem is a plus x to the power n n being a whole number this theorem is working only for whole number okay this can be written as summation of ncr a to the power n minus r x to the power r are going from zero to n this is called the binomial theorem okay or this formula can help you to expand any such binomial expression any such binomial expression okay again i would like you to understand this fact a here is the first term over here and x here is the second term don't be like be so rigid about its definition that it has to be a a means some constant or it has to be just x no i can also apply something like this let's say five x square plus one by three x to the power of 10 i can apply binomial theorem on this so this will behave like your a and this will behave like your x this is just written for you to you know keep this formula in mind don't be very very rigid on its way the way i've written it a is your first term x is your second okay one more thing i would like to highlight over here if let's say there is an expression where there's a minus sign in between okay let's say this then what do we do is we put a plus and we include the minus sign with the second term okay so same theorem will be applicable right so don't be like okay if this is plus some different theorem will be there and if there's a minus some different theorem will be there no theorem is the same is just that the minus sign is made a part of the second term okay so let's do some simple exercise before i know we talk about anything else any questions so far so far looks easy any kasta no kasta okay let's move on some simple simple question we will take and then i will complicate your life why sir all dangerous dangerous problems i'm seeing oh you easy ones okay let's do one thing uh let's ignore the question my question is just find the expansion of this term forget about the seventh term okay as of now ignore this just write the expansion of this term sir 14 terms you have to write okay let me reduce the power let me make this as uh six right i know 14 terms will be slightly too much and just say done okay you don't have to even write them on your chat box i will do it for you write down the expansion of four x minus one by two root x whole to the power of six since this is your first question take some time and write it properly done okay so first term would be by the way as i told you include negative sign with the second term okay so the first term will be six c zero four x to the power six no need to expand it please do not waste your time doing it my purpose was just to show you the expansion in general okay next term will be six c one four x to the power five minus one by two root x to the power of one okay next will be six c two four x to the power four minus one by two root x to the power of two then will be six c three four x to the power three minus one by two root x to the power of three then six c four four x to the power of two 2, minus 1 by 2 root x to the power of 4, then 6c5, 4x to the power of 1, minus 1 by 2 root x to the power of 5, and then 6c6, 4x to the power of 0, minus 1 by 2 root x to the power of 6. If you really expand it you would realize one small observation that there will be alternative plus minus terms coming up. There will be alternative plus minus terms coming up because of this negative sign. So whenever there is an odd power on this term, a negative sign will feature in between. So the presence of an alternative plus minus terms in an expansion tells you that one of the terms or the sign in between has to be negative. Is this fine? Any questions? If this is clear to you, I have a question for you. Are you second question only? This is difficult. I would like you to try this out. Even if you are not able to do it, don't worry too much about it. Find the sum of this series. Now, this is linked to whatever we had learnt a little while ago. That's why I thought I would give you these questions. So summation of minus 1 to the power r, ncr times 1 by 2r, 1 plus 3 to the power r by 2 to the power 2r, etc. Look at these terms very carefully. Up to n terms and this r is going from 0 to n. This r is going from 0 to n. Let's see how many of you are able to attempt this question. Take your time. One minute, one and a half minutes and even if you are not able to answer, don't worry. I know this is a difficult one to crack given that we just started the chapter today. Yes, any success stories to share? Nobody, not yet. Difficult to crack? Okay. See, guys, let me take this out. Let me take this out. 30 seconds, you want more. Okay, Aditya. Good. That's like a fighter. Yes, alternating sign and these terms are also written in a very peculiar way. Right, Aditya. Okay. Now, see, if you just open up this term, let's say I open up this term like this, summation r equal to 0 to n minus 1 to the power r ncr and let's take the first term 1 by 2 to the power r. Okay. Next, again, summation r equal to 0 to n minus 1 to the power r ncr. This I can write it as correct me if I'm wrong, 3 by 4 to the power r. Right, because 2 to the power 2 r is like 4 to the power r. Similarly, the third term would be summation r equal to 0 to n minus 1 to the power r ncr 7 by 8 to the power r. And this continues dot, dot, dot for m terms. So m terms are like this. Okay. Now, let us look into the first term, this term. Let's try to analyze this term a bit. If you write it down, let's say in expanded format, you would write nc0, half to the power 0. Okay. Then you'll write minus nc1, half to the power 2, oh, sorry, half to the power 1. Correct. Then again, nc2, half to the power 2. And this will go with an alternating plus minus sign till ncn, okay, minus 1 to the power ncn, half to the power n. Okay. A clear analysis of this will tell you that, sir, you're writing actually 1 minus half to the power of n expansion, isn't it? Am I not writing 1 minus or you can treat it like 1 plus minus half, take it in brackets, whole to the power n expansion? Yes or no? Does this convince you first of all? Guy 3 is convinced. Others convinced? How do you not sort the other way? Other way, is it? Oh, half minus 1, half minus 1 to the power 3. Okay. See, if it is half minus 1, of course, you would realize, see, doesn't make a difference because you're writing the same thing in a reverse way. But yes, it is very important. If you write half minus 1 to the power of n, the first term would have been nc0, half to the power n, right, which may not have a negative sign, right? This may not be the same term as this because it depends upon n actually, right? The next term would be minus nc1, half to the power n minus 1, right, which may not be the term which is just second last term. So, sign will get different over here. Aditya, sign will differ. Are you getting my point? So, till we are aware of that n, we cannot comment that it is same as, see, if n was even, it won't make a difference, whether you write 1 minus half to the power n or n minus half to the power n. Pranab is spamming something on this group. What is this Pranab? Half. What are you trying to say? I didn't get you. You're trying to say the answer. No, no, no. Answer is still to find out. Wait, don't worry. Answer is not that. Wait. So, if this is understood, can I also say that I know the second term also? The second term is actually 1 minus 3 by 4 to the power n. Third term will be 1 minus 7 by 8 to the power n and this continues till m terms. Yes or no? Since first term is, all of you are convinced that it is this, you should also be convinced that the second term is this, third term is this and so on and so forth, correct? Which means, which means you are convinced that we are trying to sum up something of this sort till m terms, okay? Not to panic looking at it because it is actually a geometric series, okay? This is actually a geometric series whose first term is 1 by 2 to the power n. Second term is the multiplication of the first with 1 by 2 to the power n. Third term is again you multiply 1 by 2 to the power n and so on. So, this is clearly a geometric series with your first term being 1 by 2 to the power n, common ratio also 1 by 2 to the power n and number of terms is m. Number of terms is m, right? So, for such a series, we already know the formula and the formula is what? What is the sum? First term, right? 1 minus the common ratio raised to the power total number of terms by 1 minus 1 by 2 to the power n, okay? This is going to be your answer. If you want to, once again, 2 to the power n, not n. If you want to simplify it, it is your call, go ahead and simplify it. But my purpose was to just convey the idea to you. Does it make sense? Was it a difficult question? Actually, no. It was based on your basic understanding of the binomial theorem structure, right? Pranav, did you get your answer now? Did you get your mistake also? Okay, right? All right. Now, we'll do some analysis on the binomial theorem that we have just now figured out, a small analysis. The first analysis is the number of terms. The number of terms in a binomial coefficient is n plus 1. But this is subject to the fact that the terms involved are dissimilar, okay? Why I'm saying you this is because, let's say, somebody says 2 plus 3 to the power of 2. You'll only get one answer. That is 25, right? So, according to this observation, it should have been 3, but the answer is only 1, right? So, this is when you are assuming that the terms involved are such that if you expand them, they don't get observed in in each other, or some of the terms do not get combined and give you a single term, okay? So, number of terms is n plus 1 under the assumption that there is no combination of like terms happening, okay? So, this is the maximum number of terms that you can get. That is what you can actually say, okay? Anyways, now, this is to be perceived as if you are trying to distribute, please note that this is actually the number of ways to distribute two identical objects, or you can say n identical objects, n identical objects between two persons, okay? If you recall this concept that we had done in our permutation combination chapter, so let's say you're trying to distribute n identical laddus to two people, okay? Let's say Sita and Geeta, and laddus have to be distributed, and these laddus are all identical. How many ways can you do that? Remember, we had learned a formula, and the formula was n plus r minus 1, cr minus 1, correct? n being the total number of objects which were identical, r being the total number of people among which we were distributing the objects, so it will be n plus 2 minus 1, c2 minus 1, which automatically comes to be n plus 1, c1, which automatically comes to be n plus 1. Now, why I'm telling you this formula is because you can scale this concept up to find out the total number of terms even if you are given a trinomial, or to be more generic even if you are given a multinomial, right? For example, let's say if somebody asks you how many terms are there in a plus b plus c whole square? I know all of you know six terms are there, right? But let's say as a person who would like to solve this from p and c point of view, I would read it as if I'm trying to distribute two laddus between three people, right? Two identical laddus, right? Or pastries, Christmas time, cakes and pastries, okay? Two identical slices of cake is to be distributed between three people. How many ways can I do that? Your answer will be n plus r minus 1, c, r minus 1, that's nothing but 4c to 6, that means there will be six terms in this expression, a square, b square, c square, 2ab, 2bc, 2ca, okay? So again, this is subject to the fact that there is no observation, what I'm saying? There's no absorption of term happening between, you know, several terms in that given expansion. Because if let's say I use this to find the expansion of this, 1x plus 1 by x plus 1. Here you will realize you only have x square, 1 by x square, 1. Then you'll have a 2, then you'll have a 2x, then you'll have a 2 by x. And now 1 and 2 can be combined to become a 3. Normally we combine the like terms, right? So there are only five terms here, 1, 2, 3, 4, 5. So this formula will fail, this formula will fail, right? So don't think that you have got a, you know, what do you call it? The all-income passing formula to solve these kind of questions. So questions may be involving some like terms also. There this method will not work or there this formula will fail, so as to say. Are you getting my point? So this is number one observation. Okay, let me ask you this question. Let me ask you this question. I think I would have a question on that. The number of terms in the expansion of this are 36. Find n. Who will tell me this? Yes, anybody? Anybody? Okay, good. Very good, Aditya. Okay, so r is 3 here. n is not known to us. But we know for the fact that n plus r minus 1, cr minus 1 is 36. We very well know the formula of ncr, which is n plus 2 factorial by 2 factorial n factorial. n factorial, this is 36. This is nothing but, this is nothing but n plus 2, n plus 1 is equal to 72, right? 72 can be written as 9 into 8. So either you compare n plus 1 with an 8 or you compare n plus 2 with a 9. Either ways you'll get n value as 7 only. Answer is 7. Okay, so if you expand x plus y to the power 7, x plus y plus d to the power 7 and x, y, z are all such terms which are dissimilar to each other such that they don't combine on expansion or they don't give like terms on expansion. Then total number of terms here will be 36 terms. Then that would be a question in itself, Aditya. So these formulas will change. So there are different ways to deal with it. Don't worry. When such problems will come, we'll basically tackle it. Okay, so yes, of course as you correctly stated that there would be certain cases given to you where combination will happen. We'll talk about it when the problem comes. Okay, don't worry about it. So now we'll continue with our analysis. So this was the first analysis. Second thing is, second thing that I would like to discuss is the general term. General term basically helps you to figure out or find out a term with a certain power of the variable. Okay, for example, there was one of the question. Remember in the first question, the question was find the coefficient of x to the power 7 in that expansion and I said, okay, ignore that x to the power 7 and just do a general expansion. Those type of questions can be solved by this concept. So listen to this. The general term is basically the one which I wrote it in the binomial theorem expansion ncr a to the power n minus r x to the power r. This is actually your r plus one-th term. Remember this. This is not your r-th term. Guys, let me tell you because r starts from zero. So r is a number which starts from zero to n. Let me write it like this. r is a whole number which belongs to zero to n. So we cannot say this is not the r-th term because if you're saying this is the r-th term means you're trying to say zero th term. Zero th term doesn't make sense. Your terms can only take natural numbers positions, right? First term, second term, third term, right? So this is an expression for the r plus one-th term. If you put r as zero, you automatically get the first term which is nc zero a to the power n x to the power zero. If you put r as one, you automatically get the second term which is nc one a to the power n minus one x to the power one. Okay. Now many people ask me, why you give this formula? Why can't you give me this formula? t r where r is your natural number from one to n plus one. See, I can give this formula to you also but you will see that this formula is much more ugly to remember. It'll look like this. nc r minus one a to the power n minus r plus one x to the power r minus one. Which is, which is the more easier formula to remember? Tell me. Isn't it this, right? So if I want, I can give you this formula also but you will not be able to remember this and a lot of mistakes will happen and in fact, you have to write more actually, right? So better to write it like this rather than writing it like this. Okay. So I will erase it in case somebody gets confused and starts using it. Is this fine? Any question? The third thing that I would like to discuss is the middle terms. Middle term bracket s because there can be either one middle term or more than one. Okay. Now, why there can be more than one middle terms? See, let me let me write some expansions for you. Let's say a plus x to the power of two. Okay. If you expand it, you get this to be more specific nc zero a to the power two x to the power zero two c one x to the power one a to the power one two c two x to the power a to the power zero x to the power two. Correct? How many middle terms do you see here? You say only one. This guy is your middle term. Okay. So only one middle term is there. However, if I write a plus x cube, you will write three c zero a cube x not three c one a square x one three c two a one x square three c three a not x three. There are two middle terms over here. See, whenever there are even number of objects, they cannot be a single middle term. There has to be multiple middle terms. Right? For example, if you have four fingers, I ask you, which is a middle finger? Right? There's no middle finger. They're two middle fingers. Okay. So these two will be middle fingers. They're two middle terms. But if they're three, then which is the middle finger this one? Okay. Sorry. Okay. Is it fine? Understood? Okay. Now, if you make a pattern out of it, tell me, first of all, whether there will be one middle term or multiple middle terms or two middle terms depends upon what you say depends upon n. Correct? So let's try to device a mechanism or a formalization of this concept. So if n is even, there will be one middle term because n plus one will be odd, total number of terms will be odd. If total number of terms are odd, they can be only one middle term. Okay. So if n is even, then there will be only one middle term. But my question to all of you is, can you tell me what is the position and the expression of that middle term? Which term will be the middle term in this case? Think and answer. You can take multiple examples, you can take a plus x square, you can take a plus x to the power four, you can take a plus x to the power six and give me a formula in terms of n as to which will be the what will be the position and the expansion or the expression for that middle term. Very good punners. Anybody else? Very good Hariharan. So yes, people are giving me absolutely correct response. It will be n by two plus one-th term that will be your middle term. Okay. Anybody who wants to contest this? Are you all convinced? You can check it out, you can see for yourself, do a bit of verification. And that expression will be nc n by two a to the power n by two x to the power n by two. Got the point. And just to remind you, this will be the greatest binomial coefficient of all the terms. This will have the greatest binomial coefficient. Guys, I'm not trying to say greatest coefficient. I will talk about it very soon in today's class only. This will have the greatest binomial coefficient. nc n by two is the greatest binomial coefficient. Okay. So the middle term or terms will always have the greatest binomial coefficient. Okay. Anyways, coming to the fact when n is odd. When n is odd, we already discussed that total number of terms would be even. If total number of terms will be even, there will be two middle terms. There will be two middle terms. Can somebody tell me what will be the position of those two middle terms? And hence what is their expression? Take example, you can start with a plus x to the power one, you can take a plus x to the power three, you can take a plus x to the power five. So take some, you know, easy to manage cases with lighter powers. n by two plus n plus half. That is three n by two plus half. Gathri didn't quite get that. n by two plus 1.5. That's a good way to state it. Okay. So here, the position of your middle terms would be n plus one by two and n plus three by two. First convince yourself. Take some examples. Okay, let's take a plus x to the power five. Right. I will not write it, but I know for sure that there are six terms into this. So t one, t two, t three, t four, t five, t six. Okay. These are the two middle terms. Now three comes from five plus one by two and four comes from five plus three by two. So this expressions are absolutely correct. Okay. Now when you have to expand it, remember we had learned an expression for r plus one is term. So you have to write it in the same way. You have to write it in a slightly lengthy fashion like this so that you know this is your r. So now you can say n c r a to the power of n minus r, which happens to be n plus one by two x to the power r, which happens to be this. Okay. This term can written as t n plus one by two plus one. So this will become n c n plus one by two a to the power n minus one by two x to the power n plus one by two. Okay. So these will be the two middle terms and also needless to say that these two will be the greatest binomial coefficient and they would be equal. They would be the greatest binomial coefficient and they would be equal. Yes, Hariharan, do you have any questions? Do you have any question to ask? Can you scroll up? Okay. How much up? This much up? Top right. Yeah, top right. Here. Any questions? Okay. Now we're ready to take up some questions. Let's start solving few questions. Let me know if everybody is done copying with this. Done. Okay. Let's have this question. Yeah, this question which we had distorted. Let's now solve this question. Find the seventh term in this expansion. Of course, you will not sit and write all the 14 terms. Can we find an easier and a convenient way to directly write down the seventh term? If yes, please do so. And just type done on your chat box so that I can discuss it. Write down the seventh term of this binomial expansion. Yes, done. Let's discuss it. Guys, this is very simple. If you want the seventh term, you have to write it like this six plus one, because then this will tell you what is the R to be used for that. So it'll be N, C, R. First term to the power of N minus R. Second term. Remember, second term will have minus inclusive. Okay. So this will give you 13 C is six, four to the power, sorry, four X to the power seven. And this you can write it as by two to the power six X to the power three. So I will not do a lot of simplification, but yes, I will do certain simplification, which will make our life easy. Two to the power eight X to the power four. Okay, no need to expand it further because you would need a calculator for it, which is not allowed anyways. Is it fine? So thanks to the general term that we are able to write such, you know, expansions without having to go to that term. Okay, the same question, the same question. If I ask you, find the seventh term from the end. Okay, let's take a second part of the question. Find the seventh term from the end, from the end in the expansion of whatever was the expansion, I think four X minus one by two root X to the power of 13. Now, I know you can do it, but the safest and the fastest way to do this problem is people start counting from backwards, which is time consuming just reverse the position of these two terms. And instead of finding seven terms from the end, find the seventh term from the beginning that will be your answer of the seven term from the end. So if you reverse the position of these two terms, then the seventh term from end in this will be the seventh term from beginning in this. Okay, and this is much easier to deal. So we can directly write it as T six plus one in this. So it'll be 13 C six minus one to the power seven, four X to the power of six. Is it fine? Okay, you can simplify this and get your job done. So if some question says find the seven term, find some nth term from the end, reverse the given binomial term positions and then write down the same position from the beginning. Okay, you don't have to reverse count it. Is it fine? Any questions? Let's take another one. Let's take this question. Find the coefficient of X to the power eight in this expansion. Done. Okay, now see, in this type of question, obviously you will not sit and write all the terms and try to consolidate the powers and see which power is giving you power of eight. Okay, so there has to be a shorter and faster way to do these kind of questions. Correct. So here, what do we do is we say, let t r plus one th term contain X to the power eight. Okay. So see, it is impossible for us to write 11 terms and then consolidate the powers of X and check which one is finally giving you X to the power eight. That would be two time consumed. Okay. So in order to do this question, you have to first assume that let r plus one th term contain X to the power eight. So r plus one th term would be what? r plus one th term will be 10 c r x square to the power of 10 minus r and minus one by X to the power r. Okay. So let's say this contains X to the power eight term. Correct. If you take all the constants together and all the powers of X together, this is what we see. Correct me if I'm wrong. Now if this contains X to the power eight term, that means it clearly implies that this power over here is actually an eight. That means three r is 12. So r is equal to four. Correct. That means your fifth term contains X to the power eight and that term itself would be 10 c four minus one to the power four. Okay. X to the power eight. So this will be called as the coefficient. Guys, again, get this clear. All the constants combined together with the variable, all the constants which is present with the variable that is called the coefficient of that variable. So this whole thing that is 10 c four, that is going to be your answer. By the way, 10 c four, I think I s 210. I have to just check it out by 24. Yeah, 210. So the answer is 210. Is this fine? Why three? Why three are here? See, we have 20 minus two are already and this X will also have an power of our right, right? Exponent rule. Exponent loss. See, I, okay. I think I have to write it in a much more easier way. See, this term 10 cr X to the power two into 10 into r is what? 20 minus two are this is minus one to the power r by X to the power r. So if you consolidate this to single power of X, what will you write? Okay, and minus one to the power r will be clubbed with the constant. This is what I wrote in a direct way. Got it. Okay. Now, one small thing I would like to add to this since we are doing these kind of problems. Let us say, let's say hypothetically, if you solve for r and r did not come out to be a positive integer. That means r came out to be negative integer or a fraction. So please note this. If r does not belong to a whole number, then the coefficient of such a term should be written as zero. That means such a term was actually not present only in the expansion. So if something is not present in the expansion, we have to write its coefficient as zero. Right. So it may happen that you're solving a question and you try to get an r and you got, let's say, a five by two or you get r as minus three hypothetically. Then basically, such a term is not there in your expansion. Okay. And if it is not there in the expansion, you can always write its coefficient as zero. Is that fine? Any questions? Any concerns? All right. We'll take more questions. Don't worry. Okay, let's take this term. If the coefficient of three consecutive terms in the expansion of one plus X to the power n is 165, 330 and 462 respectively, find the value of n, find the value of n. Coefficient of three consecutive terms. I'm not saying coefficient of first three terms, three consecutive terms. Very good, Pranav. Anybody else? Guys, do not leave the meeting. I'm taking attendance right now. Some of you have become very quiet, like Pratik. Pratik, how are you? How are you doing all this? And there is to be a gentleman by the name of Shashwat. He's not there in the class also. Who else? Rehan, Rishabh, Sauras Singh, Adansar. Yes, two people have answered so far. Others, otherwise keep participating, looks lively. Have you seen a cricket match? What is the role of a wicket keeper? His role is not only to keep the wickets. Sledging, huh? Yes, of course he sledges the opposite team, but also keeps on encouraging the team members. Isn't it? Come on, come on, you will get it. You will get it on his ball. Like that he says, isn't it? I remember they keep saying, oh, this is a weak batsman. This is a weak batsman. He'll go, he'll come and go. One ball only, he'll come and go. Rishabh Pant, I have not seen him playing, to be very, very sorry about that. Modern day cricket, I'm so much alienated from the modern day cricket. It was happening too much, IPL and all those. We used to wait for those series to happen, India versus Australia or India versus Pakistan or the Saja Cup, which is to wait for three months, four months. Now, cricket is overloaded. Some westernies, see to be very frank, I don't even know the entire team that plays for India. I know a few of them, Virat Kohli. I think I know only one person. Shikhar Dhawan, sometimes I hear his name. Not in 80s also, 80s was like beyond me. I was like, I belong to Tendulkar, Javed, sort of ganglis. Yeah, you're absolutely correct. It's not a commodity. It's no more a game. It's commodity. Yes, you can say it was an entertainment. It is always an entertainment, but not as more of a commodity. Yeah, all for commercials, yes. Some big scam will surface out after IPL and all. Same with me, yes. The first IPL I saw in 2008, and after that I never wasted my time. Because I thought it's like one of the occasions they have conducted such a gala event where all the states are participating. After that, it is every year. Who will waste time sitting in the front of TV watching 2-2 matches back-to-back? We are not that jobless also, please. Let's discuss this. See here, your coefficients will actually be the binomial coefficients. So, can I say, let's say the coefficient of the rth term, which is ncr-1 is 165. Coefficient of r plus 1th term is 330 and coefficient of r plus 2th term is 462. Now, the only way to figure out what are the values of n and r is the use of one of the properties which I discussed in the permutation combination chapter, which was this property. Ncr by ncr-1 is n minus r plus 1 by r. So, if I use it on this two, first of all, let's say, so 330 upon 165 is going to be n minus r plus 1 by r. By the way, this is 2. So, 2r is equal to n, or you can say 3r is equal to n plus 1. 3r is equal to n plus 1. This is the first formula that we get. Okay, we'll check. The next is, if you replace your r with an r plus 1, then ncr plus 1 by ncr, that will be 462 upon 330. If I'm not mistaken, this is divisible by 11. Yes or no? 11, 11, 11. Yes, 11. So, it will be 42 times 11 and this will be 30 times 11. And of course, we can go by 6 as well. So, 66 to be more, friend. So, this will be 5 and this will be 7, 7 by 5. But ncr plus 1 by ncr is just replacement of r with r plus 1. That's it. Okay. So, 7r plus 7. If you open this bracket over here, it will give you 5n minus 5r. Correct me if I'm wrong. Right? So, 12r is equal to 5n minus 7. So, clearly I can say 5n minus 7 is 4 times n plus 1. That means n is equal to 11. That's your answer. From 1 into. I mean, this was a question which was already done multiple number of times while we were doing PNC chapter, just a repetition of the same. Okay. I will take up more questions. But right now, I think most of you need a break. We started at 3.45. So, it's almost like 2 hours, 20 minutes kind of a thing. So, let's have a small break. Yeah, we definitely need a break. Okay. So, the next concept, in fact, the next problem that we're going to take is a slight complex question based on a similar concept. Oh, I think we have not done any question on middle terms. I'm so sorry about it. Let's take some middle terms question. Let's take this question. Find the middle term. In fact, they should write middle terms if at all there can be more than one. So, find the middle term or terms in the expansion of 3x minus x cube by 6 to the power 9. So, if n is odd, we have already discussed that there will be two middle terms. Correct? So, there will be two middle terms. So, what are those middle terms? Let's figure it out. Just write done once you're done. No need to simplify it. Just say done so that we can discuss it. Okay, enough. No issues. No issues. Okay. So, n is an odd number 9. Basically, your n. Okay, this is your n. If this is odd, there are two middle terms. They are two middle terms. And what are those middle terms? Tn plus 1 by 2th and Tn plus 3 by 2th. By the way, they are just one difference apart. So, if this is 5th, then this will be 6th. Okay. So, for T5, we have to use r value as 4. So, it will be 9c4 3x to the power of 5 and minus x cube by 6 to the power of 4. Okay. No need to waste time in the simplification process because this is a simple question to solve. And for 6, I have to keep my rs5. So, 9c5 3x to the power of 4 and minus x cube by 6 to the power of 5. Okay. So, these will be your two middle terms. Any questions? Easy. Okay. So, we'll start complicating it slightly. Okay. Next question for you is, which is more? 101 to the power of 50? 100 to the power of 50 plus 99 to the power of 50. Which is more? Okay. Now, of course, you have to justify your answer, whatever you are trying to, reasoning you are trying to draw from it. So, please justify your answer. Which of them is more? And please justify it. Why? Any idea anybody? Okay. 101 to the power of 50. Can I write it as 100 plus 1 to the power of 50? Okay. Can I say it is 100 to the power of 50 plus 50c1 100 to the power 49 plus 50c200 to the power 48. Till 50c50. Okay. Now, 99 to the power 50, in a similar way, can I write it as 100 minus 1 to the power 50? Now, since there's a negative sign over here, when you expand it, you will get alternating plus minus signs. Okay. As you can see here on your screen. Okay. But this will also end in 50c50. Right? Now, subtract these two. Subtract these two. Okay. Subtract these two. When you subtract these two, on the right side, if you check, you will end up getting the second term twice up, fourth term twice up like that. So, you'll get two times 50c100 to the power 49, 50c300 to the power 47, and so on. And the last term, I mean, I don't care about the last term, by the way, I don't need it. Okay. Now, try to pay attention over here. 50c1 is like 50. 50c1 is 50. Correct. Yes or no? 50 into 2 will be 100. And this is only 100 to the power 49. And other terms will be twice 50c3. I will not waste time writing it. I'll just say dot, dot, dot. Okay. Now, pay attention. Everybody please pay attention. This is 100 to the power of 50. And this is a positive term. Isn't it? This all term over here, which I'm not writing, it's actually a positive term. Right? I don't think so. There's a negativity in that term in what form, whatsoever. Right? So, what do you figure out from here? We figure out that 101 to the power 50 minus 99 to the power 50 minus 100 to the power 50 is a positive. Oh, sorry. I should have kept this symbol. Yes or no? And this is a positive term. That means 101 to the power 50 minus 99 to the power 50 minus 100 to the power 50 is greater than zero because it's a positive term. Correct? So, if this is a positive term, this is greater than zero. And if this is greater than zero, it automatically implies that 101 to the power 50 is greater than 100 to the power 50 plus 99 to the power 50. Okay? So, this guy alone is bigger than the sum of the other two. So, this is the winner. This is the term which is greater than the sum of these two. Is it fine? This is a very famous type of question. Normally, it comes in comparative exam also, but it's there in Adi Sharma also. Is it fine? Any questions? Any concerns? Please do ask. Okay. Can I move on to the next question? Let's take some more questions. Find the term independent of X in this expansion. Okay. What are the meaning of independent of X? Independent of X means it's a constant term, Gayatri. That means there's no X term related in that. That is the constant term. For example, if I say which term is independent of X in this expansion. So, when you expand it, okay, this term is independent of X. I'm just giving an example here. This term is independent of X. Independent of X means no X in that term. Right. Right. Front of when they say find the term independent of X, they do not mean the position of course only. They mean the value of the term also. This is a misinterpreted term by many students. When they see this, find the term independent, they mention the position. No. They're asking you the value of the term or the expression of the term. Very good front of. Nice. Aditya has a different answer to give. Okay, Aditya. I'm not saying right or wrong to anybody. Okay. All right. Okay, let's check. See here, when you're trying to find out the term independent of X, of course, you're not going to write all the 10 terms to see which term is independent of X. Right? It's like not a feasible way to solve the question in that way. So we'll say, okay, let t r plus one of the term be independent of X. Be independent of X. Now, we all know t r plus one of the term expansion is what n c r. This term to the power of n minus r. This term to the power of r. Okay, now consolidate all the constants together. So what are the constants present over here? Three by two to the power n minus r will be present minus one by three to the power r will be present and consolidate all x terms together. Now x terms will have a power of 18 minus two r minus r, which is 18 minus three r getting the point. Now you are claiming that this term, this guy is independent of X. Okay, if this guy is independent of X, independent of X, it means 18 minus three r should have been zero because then only your x will vanish off, isn't it? That means r has to be six. Right? If r is six means seventh term is independent of X. Okay, and that term will be nine c six three by two to the power of three minus one by three to the power of six. Okay, let's simplify this nine c six. This will give you if I'm not mistaken one by eight and will have three to the power three. Right? Yes or no? Nine c six is not much nine into eight into seven by six. Okay, and one by eight into nine nine. Sorry, nine and nine will go off eight and it will go off. So it's seven by 18. So the term independent of X over here is seven by 18. I will give you a light demonstration of this. I will use my GeoGibra tool. Sir, GeoGibra can be used here also. Yes, why not? So let's name this function as G of X. Somebody will take the custa of telling me the term three by two X square, three by two three by two X square minus minus. This is what I don't like. Minus minus one upon three X. Hold to the power off. What are the power guys? I want to go back. Okay, as you can see, okay, now since there is a nine power on the top and nine power in the denominator, they have actually taken the LCM, I believe. You'll see this term will give you a, this term will give you a term independent of X. Right? So try doing 3919104 upon 10077696. This will give you, as per our calculation, it should come out to be seven by 18. Okay, I can quickly check on my phone, does it give it like that? So 3919104 upon 10077696. Okay, this gives you 0.388 into 18 gives you seven. Yes, this is seven by 18. Clear? Any questions? Let's move on to the next question. Find the sum of all rational terms in this expansion. Find the sum of all rational terms in this expansion. Excellent. Very good. Pranav has cracked this. Well done. Anybody else who would like to contribute? No, that's not correct. Okay, let's try to look into this problem. So, see, if you write down any R plus one-th term, you will be writing 15CR, three to the power one-fifth of 15 minus R and two to the power one-third of R, isn't it? Correct? Now, if you want this term to be a rational term, that means the powers on three should be integers, isn't it? So, 15 minus R by five should be an integer. Okay, and so should be one by three R. They should also be an integer, right? So, these two should be integers, right? So, what values of R will make both of them integers, those will be the R values permissible for which I will get an R, for which I'll get the TR plus one-th term as a rational quantity. Is this point clear, first of all? Then only it makes sense to proceed further. So, does this make sense? Unless until these powers are ration, integers, I will not get a rational quantity out of it because if any one of them happens to be a non-integer, okay? Now, many times you will ask me, sir, what if three has a, you know, non-integeral power and two has a non-integeral power but it combines to give you such a power for which we get an integer, it cannot happen because three and two are prime numbers, they do not share anything in common. Are you getting my point? Okay, so now, what is R here? R has to be a number from zero to 15, right? So, let us check for which R from zero to 15 both will be integers. Let us start with this first because this looks simpler to me. So, this says if R by three has to be an integer and R has to be zero to 15, R can only take multiples of three, zero, three, six, nine, 12 and 15, right? Out of these, which will make even these as integers, let us check that also because both of them should be happy. It's not about one of them being happy, both of them should be happy. So, if you put a zero, yes, zero works for this, okay, so zero is qualified. If you put a three, you get 12 by five. 12 by five is not an integer, so three is ejected, okay? Even though it is making R by three is an integer, it is failing to make 15 minus R by five as an integer. Six, no, six will also fail, nine will also fail, 12 will also fail, 15, yes, 15 will work, okay? So, the final values of R that is going to give you a rational term is R equal to zero and R equal to 15. Now, R equal to zero is basically your first term and R equal to 15 is your last term, so only the first and the last terms were rational, rest all the terms were irrational, so all 16, out of 16, 14 of them were irrational and only two of them, that is the first and the last, they were rational. So, let's check what is the first term, so first term would be 15 C zero, three to the power one-fifth of 15 minus zero into two to the power zero. I think that gives you three to the power three, 27, 27, okay? And for 15, I will get 15 C 15, three to the power zero, two to the power five, that gives me 32 into one, so the sum will be 27 plus 32, answer is 59. Only two of you could get it, Aditya and Pranav, well done guys. No Charan, that's unfortunately not correct, but well tried, happy to see everybody trying. Go, Taitu. Yes, got it. Okay, let's take more questions. Find the term independent of X in the product of four plus X plus seven X squared times X minus three by X to the power of 11. Okay, so in this product, okay, what will be the term which will not have X in it, that is what will be the constant term that will come from the product of this? Your options are there in front of you, so I'll simply learn a poll for you guys. Let's have two, two and a half minutes for this, very good. Three of you have responded so far. Last 30 seconds I can give you. Okay, five, four, three, two, one, go. Nine of you have voted so far, what happened to the others? Is it so difficult? Okay, let's check. Out of nine of you who voted, five of you have said option B. Okay, so B and C had close calls. I see which terms will give you constants, right? So if this four multiplies with a constant over here, correct, so I need a constant in this, this X multiplies with some term containing one by X over here and the seven X squared multiplies with some term containing one by X squared over here. Okay, so what am I looking for? I'm looking for constants somewhere. Okay, I'm looking for a term containing one by X, correct, and I'm looking for a term containing one by X square because only these terms when multiplied to four X and seven X square respectively can generate a constant, right? So I will not solve this as three independent questions. I will say let R plus one term be the one which individually satisfies these criteria. So I will take first the generic version. So R plus one term of this guy will be what? 11 CR X to the power of 11 minus R minus 3 by X to the power R, right? So that gives you 11 CR minus 3 to the power R X to the power 11 minus 2R, correct me if I'm wrong, correct? Now if you want this term to be a constant, if you want this term to be a constant for constant, 11 minus 2R should be zero, which is not possible. Why not possible? Because R is going to give you 11 by 2. R cannot be a negative integer. R cannot be a fraction. R has to be a whole number, right? So there will be no constant term in the expansion of X minus 3 by X to the power 11, okay? For term containing one by X, for term containing one by X, you just have to equate this to minus one. That means 2R is equal to 12. So R is equal to 6 now. This is possible, okay? And that term would be and the term would be the seventh term of that expansion, okay? Which will be 11 C6, yeah, 11 C6 and this will be minus 3 to the power 6, okay? X to the power automatically minus one will come. You don't have to put the values, okay? So this will be your, this will be your term over here. So I would write it down over here. It's 11 C6 into 3 to the power 6, okay? Similarly, one by X square, this guy has to be minus 2. Again, this is not possible, okay? So the only term which is constant will be when your one multiplies with this. So the answer will be 11 C6 into 3 to the power 6. This is going to be your answer. Is it there in the option? Yes, option B is correct. Option B is correct. So Janta was correct. Most of you said B, however, by one more vote. Is it fine? Any questions here? Any questions? Any concerns? Please note, the constant term was not obtained and the power of 1 by X, the power of X to the power minus 2 was not obtained, right? Because our values corresponding to that was coming out to be non-whole numbers, non-whole numbers have to be rejected. Or you can say coefficient will be zero for those cases. Is that okay? Any question? Any concerns? Please do let me know. Done? Okay. Next question. Yeah. So if P to the power 4 plus Q cube is 2, P is positive, Q is positive, find the maximum value of the term independent of X in this expansion. Yes, any idea how to do this question? I think this word maximum is creating a lot of issues, right? Okay. But don't worry, there are some hints also given to you in that question. Hints. Let's see who is able to exploit that or use that. I'm launching the poll now. In case you're done with it, you can press on the poll. Good, good, good. Okay. At the count of five, we can stop the poll. Five, four, three, two, one, go. Please vote, please vote. Okay. Most of you have voted for B. That is 14 C6. Let's check. See, first of all, the process is very similar. Let TR plus 1 at term be independent of X. Okay. Be independent of X. Okay. And TR plus 1 at term here would be nothing but 14 CR, P X to the power 12th to the power of 14 minus R. Okay. 14 minus R and Q X to the power of minus 1 by 9 to the power of R. Okay. So if you want it to be independent of X, the consolidated power on, let's write it like this, and consolidated power on X will be 14 minus R by 12 minus R by 9. Correct. And this guy must be zero. This guy must be zero. Correct. Then only I'll get a term independent of X. That means 14 minus R by 12 should be equal to R by 9. Okay. So let's simplify this. 42 minus 3R is equal to 4R. R comes out to be 6. Correct. R comes out to be 6. So this term will be 14 C6, P to the power 8, Q to the power 6. Correct. Now the problem is I have to find the max value of this subject to the fact that P 4 X, Q, Q is 2 and P and Q are positive terms. Guys, this is a big indication to you that you have to use AMGM inequality. You have to use AMGM inequality. I'm getting the point. So can I say if P and Q both are positive, so will be P to the power 4 and Q to the power 3. Correct. So can I say P to the power 4 plus Q to the power 3 by 2 would be greater than equal to P to the power 4, Q to the power 3 under root. Correct. Yes or no. That means 2 by 2 is greater than equal to P to the power 4, Q to the power 3. That means P to the power 4, Q to the power 3 is less than equal to 1 and so is P to the power 8, Q to the power 6. That means max value of this fellow is going to be 1. So max value of this is going to be 1. That means max value of this entire thing is going to be 14 C6. Answer is option number B. Janta is absolutely correct. Now people have solved it. I'm sure you are not aware of this fact that you can use AMGM inequality. Is it fine Pranav? Do you want to copy from here, this page? Then question. Do let me know once you're done. Correct. Next concept that we are going to talk about.