 Alright, so we left off the last class discussing finding the derivatives of functions through first principles and today I'm planning to do a bit more. I'm planning to start with methods of differentiation along with few illustrations on it and then I'll switch on to permutation and combination chapter. Okay, so let me begin with methods of differentiation. I know you are all aware of it but officially we have not done this after the bridge course. So just a quick revision for those who have already done this. Methods of differentiation, we also call it as M-O-D, M-O-D. Anurag, you have to click on the call from the devices from your phone if you are from the phone. If you are attending it from the phone, please click on the audio button of that app. When you join in the session, it would have come. Okay, great. Alright, so method of differentiation, the first method that we are going to talk about is when you have a function multiplied to a constant K and you're differentiating it, okay, it is as good as K times the derivative of the function. Okay, for the purpose of writing less, I will refer f of x as f. Okay, hope everybody is able to hear me properly. Please check your audio settings, RMR from your phone if you are from your phone or laptop. If you are from the laptop, click on computer audio. Is it fine now, RMR? Haven't you all done a session on zoom before? Then how come you are not able to find the proper settings? I think at least two, three classes have happened in past on zoom. Let me know if you are able to hear me now. RMR, able to hear? Yeah. Yeah, I can hear you also. Okay, what about analog? Okay, man, are you all able to hear me? Okay, thank you. Thank you. Thank you. Okay, fine. Chalo, we'll begin with the session. Okay, so our second property is the sum or difference rule. Okay, sum or difference rule. It says that if you are differentiating the sum or difference of two functions with respect to x, it is as good as summing or finding the difference of their respective derivatives. Okay, this is called the sum slash difference rule. Okay, these rules we have already done in the past, so I'm not going to spend a lot of time on them. Next is the product rule. Product rule is where you have multiplied two or more functions and you are trying to find out the derivative of the product. Correct? So the rule says that you have to differentiate the function keeping one as such multiplied with the derivative of the other one at a time. So for example, here I have kept f as such multiplied with the derivative of g, then g as such and multiplied with the derivative of f. This is called the product rule. And I'd also discussed with you that if there are more than two function involved, right, how does this pattern work? So let's say if there are three functions involved f, g, h. So what do you do? Again, you do the same thing. You differentiate one function at a time keeping the other two as such. For example, you can do f, g as such derivative of h with respect to x. Then you keep f as such, f and h as such derivative of g with respect to x and then keep g and h as such derivative of f with respect to x. Okay. In short, you can remember this like this. uv dash is uv dash plus u dash v. Okay. If you are more than, let's say there are three functions, then it's uvw dash, uv dash w, u dash vw. Remember one at a time keeping the other two as same. So if let's say if you have three functions, uv, w and t, then you will do uv, w, t dash, uv, w dash, t, uv dash, w, t, then u dash v, w, t. Okay. So one at a time keeping the others intact. Okay. This is the product rule. Okay. Fourth rule that we are going to talk about is the quotient rule. So if you are finding the derivative of, if you are finding the derivative of f of x divided by g of x, that is the quotient of f and g. So the rule says it's g derivative of f minus f derivative of g whole divided by g square. Please ensure you have taken the right order. Okay. So please ensure the order is taken properly. Ensure you have taken the order of the numerator properly. Okay. So I'm writing down over here. Many of the students who are mistake, they do fg dash or f into derivative of g minus g into derivative of f. That would be wrong. That would give you the negative of the actual answer. So ensure you have written the order in this way. gf dash minus fg dash by g square. So let me write it in terms of u v also. So it's v u dash minus u v dash by v square. Okay. This rule is called as the quotient rule. So in future, yes. But could we also write it as d by dx into f into 1 by g and use product rule. But when you're using product rule on 1 by g, you have to use chain rule there. Yeah. Okay. So you can't do that but use chain rule. Of course, you can do that very well. You can do that. But you have to use chain rule on while differentiating 1 by g. Yeah. Okay. So the last tool, which I'm going to take in the next page is the chain rule. Okay. Now chain rule is applied to composite functions. Chain rule is applied to composite functions. What is composite functions? When you have something like a function fed as an input to another function. So let's say I have a function e to the power x. I have another function tan x. Okay. So if I make a composition of a function like this, this is called a composite function. Also called as f o g. Also known as f o g. Okay. Many people write it like this. This is to be read as f of g of x. f of g of x. Okay. Many people call it fog and all those things. But that's just a slang way of representing it. But it's f of g of x. Okay. So what does it mean? You are you are feeding tan x as input to this function. So in place of x, you are feeding tan x. So it becomes e to the power tan x. Okay. Now this rule says if you want to differentiate a composite function, so if you want to differentiate a composite function, let me write it like this f of g of x. Okay. This rule says then you have to first differentiate f with respect to g. You have to first differentiate f of g of x with respect to g into derivative of g with respect to x. Please note, it creates the same impact as if dg and dg are getting cancelled from the numerator and denominator over here, leaving you with d by dx of f of g. Okay. Many a times, we don't want to write this d by dx. So I have seen students using this expression f f of g of x whole dash is nothing but f dash g of x into g dash x. Okay. By the way, let me tell you here a lot of students are still not clear with these representations. Okay. The other day I asked a student what is the difference between what is the difference between f of x cube dash and f dash x cube. And he said both means the same thing. Does it? I want to ask you all, does it mean the same thing? If not, anybody can unmute yourself and explain me how are these two expressions actually different? Anyone or even you think both are the same things? Okay. Let me clarify over here. This means you are finding the derivative of f of x cube with respect to x. Okay. This is the meaning of this. Whereas this means you are finding the derivative of f of x cube with respect to x cube, my dear students. Let me tell you. So whatever function is inside, you are actually finding the derivative with respect to that function. Okay. So both are not the same things. Be very, very careful while you are reading these type of expressions in your question paper, whether it's cool exam or J main exam. Correct. So see here, this expression and this expression, there's different expressions. Okay. Had they been the same expressions, I would have just equated them, but they are not. But they are not. Okay. So I'll give you an example. So let's say if you want to find the derivative of e to the power tan x, okay, with respect to x. So this guy says you have to differentiate e to the power tan x with respect to tan x with respect to tan x, then multiply the derivative of tan x with respect to x. Are you getting it? Okay. So now who will tell me what is the derivative of e to the power tan x with respect to tan x? So if you assume your tan x to be like you. So basically what you are finding here, what are you finding here? You're finding the derivative of e to the power u with respect to u, isn't it? That is nothing but e to the power u only. So this term will come in place of this answer. And this answer we already know is equal to secant square x. So the answer for this will be, I'm writing down over here, the answer for this will be e to the power u, u is tan x into secant square x into secant square x. Are you getting this point? Okay. So my objective here is to explain you the difference between these two terms. Please don't don't get confused between these two terms. Okay, they're different. Let me cite another example. Let's say I want to differentiate under root of sin x. Okay, I want to differentiate under root of sin x. Let me tell you this is also a composite function. Okay, how is under root sin x formed by using two different functions? One is root x function and other is sin x function. Yes or no? Do you agree with me? So if I do f of g of x in this case, it will become under root, put the sin x in place of x, put the sin x in place of x. This is how a root sin x function is formed and it is a composite function. So please remember this. This is a composite function. Okay. Almost 90% of the cases that you would find in your school exam or in normal calculus would involve some kind of composite function. Okay. Now, how do we differentiate this simple? The rule says you first have to differentiate root of sin x with respect to, with respect to g, g sin x itself. Are you getting it? And then you have to differentiate g function with respect to x. Is that clear? Just recall this formula. If you want to differentiate f of g of x, you first have to differentiate f of g with respect to g, with respect to g, with respect to the inside function and multiply it with the derivative of g with respect to x. Yes, Arpita, you want to ask something? Arpita, you raise your hand. Any question, guys? Please feel free to stop me. Please unmute yourself and please feel free to interact. Okay. So let me complete this now. So what are the derivative of root sin x with respect to sin x? So it is like, let's say I take sin x as u. Aditya, you can watch the recordings. I cannot repeat whatever was done till now. We were just doing the methods of differentiation. Arpita, you check on your, even Arpita is not able to hear me. Yeah. So this is nothing but you are finding the derivative of root u with respect to u. Correct? We all know that derivative of root of something is derivative of this with respect to u, u is power rule of differentiation. So half u to the power half minus one. That's nothing but one by two root u. Okay. So put your u back as, put your u back as sin x. So we say derivative of root sin x by with respect to sin x is going to be one by two root sin x. Okay. So this term will become one by two root sin x into derivative of sin x. We all know as cos x. Okay. So this becomes the answer. This is the chain rule. Can somebody type for Arpita what to do on the screen because she is unable to hear me. Yeah. Arpita, are you able to hear me? You can turn off your camera. Just check on your audio settings, computer audio settings. Arpita, are you able to hear me? Okay. Just give me a second guys. Fine. So now let me tell you something very interesting over here. Normally chain rule is even used when you're trying to differentiate a function which is in terms of some other variable and you're differentiating with respect to some other variable. For example, if I asked you what are the derivative of y square. Okay. Of course, y is a variable over here. y is a variable. What are the derivative of y square with respect to x? Can anybody tell me the answer for that? Unmute yourself. You can talk if you want to speak out. So 2y into dy by dx. Absolutely. And you know what you're using there. You're actually using a chain rule there. See, first you're doing derivative of y square with respect to y and then doing into dy by dx. It is not 2y, that's the point. It is not 2y, my dear. It is going to be 2y and this additional term will come out, dy by dx. If you miss out on this term, everything will go wrong. Are you getting my point? So even for those cases where you are trying to differentiate a function which is in some other variable, for example, here in this case, my function is y square. It is in terms of y, but I'm differentiating with respect to x. Then also chain rule is helpful. Then also chain rule is helpful. Chain rule is so important for our differentiation that almost everywhere it is finding its application. Quickly, I would like to ask you, what is the derivative of sine t with respect to x, where t is a variable? What would you say? Cos t into dt by dx. Cos t into dt by dx. Wonderful, wonderful, wonderful. Cos t into dt by dx. Must not forget this term. Very important. Now, what I'm going to do is, I'm going to test you with a slightly difficult question. Meanwhile, one example I will take for you. Let's say I want to differentiate tan to the power of 3 of y cube. And I'm differentiating this with respect to x. All of you, please note the question says you are differentiating tan cube of y cube with respect to x. How will I do this? Now, I know most of you would hate this method because it involves writing too many things. Even I do not recommend using this method for practical problem solving. So, there's an approach which I normally tell and if you follow that, you'll be able to get all your answers. If you recall from your bish course, what I used to say, think as if there's a treasure at y. There's a treasure at y and let's say you are a thief and you want to reach to that treasure. Now, you have been stopped by various gates. Now, these gates are the functions that you meet while you are coming towards y. For example, if I see tan y cube is actually tan y, tan cube y cube is actually tan y cube whole to the power 3. So, let's say we want to steal the treasure at y. So, what we'll do, we'll start from the outer gate. This is the outer gate. So, read this as something to the power of 3. So, first, identify which is the outer most function in that composite function and read everything inside it as something. So, something cube you would say this. So, please pronounce it in your mind. Something cube. Something cube derivative will be 3 something square. 3 something square. Getting my point. 3 something square. Has Arpita joined in? No, she hasn't joined yet. Next, once you have crossed this cube gate, once you have crossed this cube gate, think as if this is not there anymore. Now, next you see tan something. Next gate is the tan gate. Tan something. Tan something is seek square something. Okay. Please keep noting down that something. Something will change every time. Are you getting the point? Next, when you have crossed tan, erase tan from your mind. Now, you have y cube. y cube derivative you can also do without any gate thingy. You can say y cube derivative. Just now, you told me y square derivative is 2y dy by dx. So, y cube derivative will be 3y square dy by dx. Now, once this is also done, everything has been crossed means your task is done. So, this is going to be your answer. Is that fine? Any doubt regarding this? Any questions? Please type CLR on your chat box if it is clear. Type CLR on your chat box if it is clear. Awesome. Now, I'm going to take some few problems. I'll not take much of your time because these concepts are quite easy. So, let me begin with this question. Hope you can read the question here. The question is f of x is defined as log of log x and the base of the outer log is x. The base of the inner log is e. Dear students, please note if there is no base mentioned with a log in calculus, we take it as a e. If no base is mentioned, we take it as e. Okay. Let's see who is able to find this. Please type in your response in the chat box. The question is finding the derivative of log of log x to the base of x at x equal to e. It's a direct application of your chain rule that you have learned just now. Arpita, are you able to hear now? Arpita, log x by x square. No. See, you have to put the value of x as e after you have differentiated. What are the meaning of this term? The meaning of this term is after you have found out the derivative of the function, put x value as e. Okay. You'll get an answer. So, is it 1 by e square? 1 by e square is wrong. Sir, 1 by e. 1 by e is absolutely correct. Who was that? Siddharth. Siddharth. Very good. Others, please give it an attempt. So, 1 by e is the right answer. Sir, she said her net is very slow. So, she's not able to open. It doesn't depend on net though actually. I mean, if the net is slow, my voice will crack, but at least she will be able to hear me. Dhyan says, hey, Dhyan, you need to put the value of x as e. Okay, let's discuss this. Let's discuss this. Now, first of all, I would like you to pay attention to the fact that you can't differentiate with respect to x being the base of log over here. So, you first have to use the change of base property, something like this. Okay. This is called the change of base property. Change of base property. Okay. So, what I did, change of base property, if you would all recall, log m to the base n could be written as log m to the base b by log n to the base b. Okay. So, I converted it to the base of e because we know the derivative of log x with respect to e. What is that? Guys, hope you are aware of all the formulas that are required for differentiation. If not, I'll just quickly note it down. So, list of the formulas. So, please quickly make a note of it. I'm going to be slightly fast. Derivative of a constant is 0. Derivative of x to the power n is n x to the power n minus 1. Derivative of sin x is cos x. Derivative of cos x is minus sin x. Derivative of tan x is secant square x. Derivative of cosec x is minus cosec x cot x. Okay. I'll write it here also. Derivative of sec x is sec x tan x. Derivative of cot x is minus cosec x square x. Derivative of e to the power x is e to the power x. Derivative of log x to the base e, that is only 1 by x. Okay. If you have some other base, you have to change the base of your log and then do it. Okay. If you have a to the power x, where a is some positive number, it is a to the power x log a to the base e. Okay. Apart from that, you need to know the derivative of inverse trig functions also. So, I'll give you that inverse trig functions because we may need it today. So, So, can you go back to the previous slide for a second? Anjali, can I just complete this and then go? Yes, sir. Okay. Thank you. So, sin inverse x, it would be 1 by under root 1 minus x square. I've got it. It is my face scene. Is your face scene? No, no, no. It's not scene. Okay. Just don't open your webcam. Just, keep the volume up. Okay. Arpita? Yes, sir. Thank you. Okay. Welcome. Cos inverse x is minus under root 1 minus x square. Guys, note this down because you may need it. You may need this in problem solving. Derivative of tan inverse x is 1 plus x square. Derivative of cot inverse x is minus 1 plus x square. Derivative of seek inverse x is 1 by x under root x square minus 1. Whereas derivative of cos inverse x is negative 1 by x under root x square minus 1. Okay. These yellow ones are the ones which you may not have done in school, but just keep a track of it. Under no situation, anybody should forget these formulas. These formulas must be internalized. Sir, can you move right towards the inverse trig functions? I'm sorry. This is important log x derivative with respect to e with respect to e, sorry, log x with the base of e that only can be written as 1 by x. If let's say somebody gives you find the derivative of log x to the base of a where a is not e, then how would you do it? You will say let me first change the base to e. So first you would do this. Okay. Remember, log a to the base e is just like a constant. So this guy is just like a constant. Okay. So you will say something like this. 1 by log a to the base e derivative of log x to the base e or ln x. So your answer will be 1 by log a to the base e into 1 by x. So this becomes your answer for those logs where the base is not e. Can I now go back to the problem everyone? Sir, can you go to the first table just one second? First, this one? Yeah. Sir, can you derive inverse derivative of inverse trig functions using first principle? I can derive, but a bit of property of inverse is required. So let me go back now to the previous board where we were talking about this question. Okay. Now think as if it is u by v type. Basically, you have to start with a quotient rule. Let me tell you all these rules, they work in tandem to calculate the derivative of a function. It's not like the entire problem can be solved just by a product rule or just by a quotient rule or just by a chain rule. You may have to use one within the other depending upon the type of questions being given to you. So here the broader structure is there has been something divided by something. So this is let's say u, this is let's say v. So it is a u by v type. Okay. So according to the formula, the derivative of this would be what? Derivative is v u dash minus u v dash by v squared. So just write a broad framework, right? So overall broadly, you are applying your quotient rule. Now within that I'm going to fill in the required expressions. So your v is nothing but log x. Okay. Let me write it as ln x in order to save my time. Okay. Now u dash, this is your you my dear students. So if you want to differentiate this, you have to use chain rule, right? So read this as ln something first. Okay. Put the brackets. Okay. Don't do conjuci inputting brackets. Don't do conjuci inputting brackets. It is for your own good. Somebody's mouth is on. Right. One, one. So u dash would be what the derivative of ln of ln x. So ln something first you read. ln something is one by something into. So forget this ln from your mind. You have crossed the first gate, next is ln x itself which is again 1 by x. Are you getting this point? Okay. Into, sorry, back it close, minus u, u is the upper function which is ln of ln x into derivative of v, derivative of this is again 1 by x. Okay, again, put proper brackets. The brackets are there to help you not to get confused. Divided by v square, v square is ln x. Since I have used ln everywhere, let me use ln only, ln x square. Now, this is your derivative. Now, the question is asking me, what is the derivative at e? At e means wherever there is x, you have to substitute e. Sometimes they would write this as f dash e also. Guys, remember, when they say f dash e, it doesn't mean you have to put e first and then differentiate. Then you'll get a zero as your answer every time, right? So, what they mean is first you differentiate and then put x as e. So, be careful. So, learn these notations because if you don't learn them now, you'll never be able to connect to these concepts later on. Now, ln e is 1. So, I'm just starting putting a value of x as e. So, 1 by ln e is again 1 by 1. Again, 1 by e minus ln of ln e is ln 1 into 1 by e. By ln e, ln e is again 1. So, let me write 1 only, 1 square. So, if you simplify this, it just becomes 1 by e because ln of 1 is 0. This guy is 0. So, everything will here become 0. So, you're only left with 1 by e divided by 1 square. Is that clear? Is that clear to everyone? Any questions regarding this? If there is no question, please type NOQ. No questions. NOQ on your chat box. Everyone. Okay. Okay. Great. Great. So, when it is clear, please type CLR, when there's no question, NOQ. Only few of you have typed. Anyways, let's move on to the next page. Sorry? Sir, in page 7, that log derivative, could you show that for me? Oh, okay. This one? Yes, sir. Is that fine? Sir, can't see. Can't see the one below, the one below. This one? Yes, sir. Okay. Is that fine? Yes, sir. Okay. So, now let me move on to the next page. If y is equal to sin inverse x by under root 1 minus x square, then 1 minus x square dy by dx is equal to option A is x plus, sorry, option 1 is x plus y, option 2 is 1 plus x y, option 3 is 1 minus x y and option 4 is x y minus 2. Just two minutes to solve this, not more than that. I'm going to watch out on the time taken. Once done, please type in your response, option 1, 2, 3 or 4 on the chat box. Derivative of sin inverse x, we all have seen just now in the formula list is 1 by under root 1 minus x square. I'll write it down over here. Derivative of sin inverse x is 1 by under root 1 minus x square. Okay. Ananya R has responded, Shisi has responded, Vibhav has responded, Kirtana has responded, good guys. So, I'm getting two types of answers. One is option 2, another is option 3. Okay. Okay, fine. Let's quickly check into what is the actual solution for this. So, first of all, this expression that is y is equal to sin inverse under root 1 minus x square. I'll take this denominator on the other side. I will write it like this. Okay. Okay, no problem with this. Now, I will differentiate both sides with respect to x. So, let me differentiate both sides with respect to x. Now, while I'm differentiating my left hand side, please note that there are two functions here. One is in terms of x, other is in terms of y. So, I'll be using my product rule for this. Okay. So, first I will keep under root 1 minus x square as such. What is the derivative of y with respect to x? Remember, derivative of y with respect to x is dy by dx. Okay, don't write 1, right? It is not x, right? Had it been x, then derivative of x with respect to x would have been 1, but this time it is a different variable y and you are differentiating with respect to x. So, you should say dy by dx. Okay, plus keep y as such. Now, for finding the derivative of under root 1 minus x square, you will have to use chain rule. So, read this as something to the power of half. Something to the power of half. Okay, because half is the outer function. Something to the power half is half. Something to the power half minus 1. Okay, into. Now, forget this under root sign from your mind. What is the derivative of 1 minus x square? You'll say it is 0 minus 2x. Guys, everyone is clear with this step because this problem is actually trying to test you on various fronts. We started with the product rule. Now, we are inching towards the use of chain rule within that. Is it clear? Everyone. Okay, no problem with that. Anyways, is equal to sine inverse x. Sine inverse x is nothing but 1 by under root 1 minus x square. Okay, let us try to bring this term in the shape because right now we have written too many expressions. Let me simplify them. 1 minus x square dy by dx. I have copied this as such. This will be y by 2 under root 1 minus x square into minus 2x. Okay, is equal to 1 by under root 1 minus x square. So this 2 and this 2 gets cancelled. Let us multiply throughout. Let me write it. Let us multiply throughout by root 1 minus x square. Okay, so everywhere I'm multiplying with root 1 minus x so this will become 1 minus x square dy by dx. This will become minus xy. This will become a 1. Correct. So this term is 1 plus xy. Is that fine? So which option matches with my given scenario? Let's check. Which option matches option 2 matches option 2 matches. So guys, option 2 is the right answer. Option 2 is the right answer. Okay, well done to all of you who got option 2 as your answer. Very well done. We'll take one more question quickly. Hope you can read the question. Y under root x square plus 1 is equal to log. This log is to the base e when no base is mentioned. Remember, I told you it is taken to be as base of e. Then you have to show guys these type of questions are very important in your school. You would realize that such questions are asked a lot. So again, your time starts now two minutes to answer this not more than that. Yes, yes, yes, these slides will be shared to you. The video will also be shared to you. Prajna, don't worry about that. Every online session, whatever happens, the video will be shared to you after the class so that you can watch it later at your own pace. It will be shared to you on the YouTube. I'll just give you another half a minute in the school and have they discussed about double derivatives by any chance d2 y by dx square. No, sir. They totally can't use it. Okay. Anyways, that falls in the higher order derivatives, which is a part of your class 12 syllabus. But many of times I've seen in school, they do it little early so that, you know, you understand those double derivative concepts in class 12 very well. Anyways, we'll not do it if it is not done in your school, but after the exams, we will do these chapters in much more detail. Okay, nobody has been able to solve this till now. Okay, so let me help you out with this. So again, your expression is y under root x square plus one is equal to ln of under root x square plus one minus x. Okay. Now, let me differentiate both sides with respect to x. Let me differentiate both sides with respect to x. So while differentiating the left hand side, I'm going to use product rule. So first, why I'm going to keep as such. Derivative of under root of x square plus one. Which rule will I follow for this? The chain rule. Correct. So first read this as something to the power of half. So half something to the power of half minus one into forget the root from top. Now you have crossed that root gate. Next you see x square plus one that is going to be two x plus zero. Okay. Please note x square plus one has to be differentiated at the same time. Okay x square plus one is not a composite function. It's just an addition of two functions x square and one. So you have to differentiate it at the same time. Okay. Now this is that next keep the x square plus one term as such. Derivative of y with respect to x will be d y by dx derivative of y with respect to x will be d y by dx. Okay. So trippin says zero. Let's let's check it out. And then you also say zero. Okay. Now what about the derivative of the right side? Ellen something again. There's a chain rule here. Ellen something Ellen something is one by something. One by something. Okay. Put proper brackets brackets will always help you out. Now once Ellen gate is gone. There is two functions over here. So first function again. You have to use a chain rule. Okay something to the power half half something to the power half minus one. Okay into derivative of x square plus one is 2x plus zero. Okay. Minus derivative of x which is going to be minus one. Are you getting this point at any stage if you are missing out any kind of a derivative your answer will go wrong. Please ensure through these examples you have understood this process very well. Correct now time to simplify. Here if you see you can write it as y by two under root x square plus one into 2x. This is nothing but under root of x square plus one d y by dx. Let me simplify this as well into this term again is 2x by 2 under root of x square plus one minus one. Okay. So this to this to goes off here this to and this to goes off. So we'll be left with x y by under root x square plus one. This will be under root of x square plus one d y by dx and this will be one by under root x square plus one minus x here if you take the LCM. It becomes x minus under root x square plus one is that fine. Now I'm sure you would have recognized by this time that this term is the negative of this term. This term is the negative of this term. Okay. So what I'll do I'll cancel it off and put a minus one over here. So ultimately you get x y under root x square plus one plus under root x square plus one d y by dx. Is equal to negative one by under root x square plus one. Now both sides you multiply by under root x square plus one. Okay. Both sides you multiply with under root x square plus one that will give you x y. So this term will go off. This will become a whole square. So x square plus one d y by dx. This term will become a minus one. So ultimately you end up getting that expression that you desired by taking the plus one on this side. So the right hand side is actually a zero. That is option number one is correct. Is that fine? Anybody having any questions anywhere you have not understood especially the use of the chain rule. Please unmute yourself and ask the questions. No questions. If no questions then we'll move on to the next one. This is the last question that we're going to solve and then we're going to start with permutations and combinations. So it's given that cos y is equal to x times cos a plus y. Okay. Prove that d y by dx is cos square a plus y by sine a. Two minutes for this guys. Your time starts now. Just type done on your chat box once you're done. Since it is approved that question. You cannot tell any options. Just say done if you're done anyone done. Two minutes already over. Okay. Let's look into this. First, let's write x as cos y. Oh, she's done very good. So first, let us write x as from this expression. As you can see from here, you can write your x as cos y by cos a plus y. Okay. No doubt about it. Now, let us differentiate both sides with respect to y with respect to y. Remember now we are differentiating with respect to y not x. Correct. So what are the derivative of the left hand side, which is x with respect to y? You will say simple. It is dx by dy right. What are the derivative of right hand side with respect to y? You say I have to use quotient rule over here. So you say v u dash u dash u dash means derivative of cos y, which is minus sine y minus cos y into derivative of cos a plus y, which is minus sine a plus y. Okay. Please note a plus y derivative will be zero plus one. So you don't have to write it because that will anyways be one actually. Okay. Whole divided by cos square a plus y guys. Remember, I'm not writing dy by dx anywhere because I am differentiating with respect to y not x. So it's very important to realize with respect to what are you differentiating the function with? Okay. Are you differentiating with respect to x? Are you differentiating with respect to y? If you're differentiating something with respect to y and the function is already in terms of y, you don't have to put a dy by dx etc anywhere. But if your function is in terms of x and you are differentiating in y, you have to put dx by dy terms like that. Okay. So it's just a bit careful use of your chain rule be very very careful. Now if you expand this, it becomes minus sine y cos a plus y and this will become plus cos y sine a plus y by cos square a plus y. I'm sure most of you would have identified by this time that if you take a minus sign common from here or if you just swap the positions of these two terms, it is actually a compound angle formula. So it is something like this sine a plus y cos y minus sine y cos a plus y by cos square a plus y. Now this is just like reading sine a cos b minus sine b cos a. Okay. Just compare it with this formula. So this term is basically using a compound angle identity which is nothing but sine a minus b. Sine a minus b is this. Okay. Divided by cos square a plus y. This is your dx by dy. My dear students, let me tell you. Okay. Now what do we have to find out? We have to find out dy by dx means we have to reciprocate this. We have to reciprocate this. This becomes your desired result. Is it what we wanted to prove? Oh, yes. This is what we wanted to prove cos square a plus y by sine a. Okay. Couple of things that come out as a learning from here is that dy by dx. Let me write it as learnings dy by dx is reciprocal of dx by dy. Okay. dy by dx is reciprocal of dx by dy. But let me tell you d2 y by dx square which you learn later on is not the reciprocal of d2 x by dy square. Okay. This is not true. Okay. This is not true. Okay. Don't use this very important. But this is true dy by dx and dx by dy are reciprocals of one another. Okay. So guys will stop here and again, I will sincerely request you do your n crt problems of your limits derivatives. In fact, all the chapters that are going to come. Okay. So I'm not going to begin with permutations and combinations permutations and combinations. Now this chapter normally I say it comes directly from God's book. Why? Because it is basically dependent on your art of counting and actually I call it as an art is just like coding. You know, when you're asked to solve a problem by writing a code, some of you may write very few lines and solve the problem and some of you may take a longer approach to solve it. Okay. It's more of an art. I would say some people are inherently blessed with this art, but some people who are not we need to practice and we need to expertise in that through a lot of practice. Okay. So even I start talking about what is permutation? What is combination? Okay. I'll talk about the prerequisites for this chapter or some important tool. The requisites or some tools that we need to learn. Okay. These tools will help us to count. Okay. These tools will help us to count. Let me tell you the importance of this chapter. This chapter will be directly influencing your probability chapter as well. Okay. Not only in class 11th, but also in class 12th and probably in your first year of undergrad as well. So so important is this chapter. So if you're not confident about PNC, please ensure you have practice enough to make yourself confident in it. Else what will happen? You will have to you will feel lack of confidence in probability chapter as well. Okay. So two chapters will, you know, get influenced by this chapter. So let us start with the prerequisites. The first thing that we need to talk about here is the factorial representation. I'm sure you would have done factorials in school. Okay. What is factorial? First of all, we represent a factorial of a number n where n should belong to a whole number. Okay. But the use of n followed by an exclamation mark. Okay. What is this notation actually? How is it defined factorial is defined as product of n n minus one n minus two till we reach till we reach one. Okay. So for example, if I say five factorial, it means five into four into three into two into one, right, which is actually 120. Okay. If I say four factorial, it means four into three into two into one. Right. That is 24. If I say three factorial three into two into one. Okay. So basically it is just a shortcut notation which will make your computation of the product of such numbers starting from n up till one a simple looking expression. So two factorial. What is two factorial two into one? Right, which is two one factorial one, which is one. Now there's something of zero factorial also, right? Zero factorial is also one. Now you must be wondering how yes or no, you must be wondering how. Okay. Now the answer to this is hidden in the properties of factorial. Now if you would see this property carefully, if you were one, if you wanted to get one, two factorial, it is nothing but one factorial multiplied with two. Isn't it? Okay. If you want to get three factorial, that is six, you just multiply this number with three, which is nothing but three into two factorial. Correct. If you want to get to four factorial, you just multiply four to six. Isn't it? So it is four into three factorial. If you continue this trend, if you continue this trend, you realize any factorial n could actually be obtained by multiplying n with the factorial of the previous number. Right. n factorial is nothing but n into n minus one factorial. That is most important property of factorial. This property will actually help us to write a higher factorial in terms of a lower factorial or a higher number factorial in terms of a lower number factorial because you can scale this up. So if you want to write it in terms of n minus two factorial, you can write it as n, n minus one, n minus two factorial. So you could represent n factorial in terms of n minus two factorial. Okay. If you still want to go down, you can do it n, n minus one, n minus two, n minus three factorial, etc. Are you getting my point? So this helps us to express a higher number factorial in terms of a lower number factorial. Okay. Now, how does this help us to prove zero factorial is one? Let's take it out. So in this formula, that is n factorial is equal to n, n minus one factorial. I'm going to put n as one. Okay. When you put this, you get left hand side as one factorial. Right hand side is one. This becomes zero factorial. Now, zero factorial is something which we don't know, but we know the value of one factorial and one, which is nothing but one itself. So this brings me to the fact that zero factorial is also one. Is that fine? Now, couple of problems we'll take on factorial and then we'll see where all factorials are helpful to us. Okay, especially one concept I'm going to take up today, which is not there in your ncrt. So, uh, let me just go to the next page first. Yeah, a very simple problem to start with. Find n if n plus two factorial is 60 times n minus one factorial. Give me the value of n two minutes for this time starts now. So three awesome. Very good. Who was that? Are the other very fast. Good. Three, three, three. Okay. People have started responding. Ajay, Ronak, Vibhav, well done. Well done. All of you good. I think you were well within half a minute. So since you see guys, let's solve this. It's a simple problem. Since you see n minus one factorial appearing. Okay. I would try to write a bigger factorial in terms of the lower factorial so that we can cancel them out. You know, both the sides. Remember none of the factorials is zero. So cancellation can be done on both the sides. Okay. Remember a non zero term only can be cancelled from both the sides of the equation, not zeros. So n plus two factorial can be written as n plus two n plus one n into n minus one factorial. I've stopped here. Why? Because my right hand side also contains n minus one factorial. So I will write it only to that factorial where I need it. Okay. So this is equal to 60 times n minus one factorial cancel them off. So basically you are left with n plus two n plus one into n is equal to 60 guys. Let me tell you most of the students make a cubic equation out of it and try to solve it. Remember that would be a lot of waste of time. Remember here n is a whole number. Okay. So think of three consecutive whole numbers which multiplied to 60. It's obvious that it is five into four into three. Isn't it? Okay. So you can make this assumption. Okay. And break 60 as five into four into three, which means either your n plus two is five or n plus one is four or your n is three. All of you give you all of you give you the same result for n which is three in this case. Okay. Hope you have not guessed it guys. Hope you have not guessed it. Okay. Next question. Okay. Next question. I'll give you from my end. Find the remainder. Find the remainder when when summation of our factorial are from one to 100 is divided by is divided by 15. You have to give me the remainder when summation of our factorial from R equal to 1 to 100 is divided by 15. Let's say let's see who gets these answer fast time starts now. Okay. Five. Shomik says three. Well, I'm getting different answers. Let's wait for at least three more people to respond. Okay. Rajna also says three. Aditya Shinoy also says three. Okay. Let's let's discuss it guys. So first of all summation of our factorial. Most of you are saying three. Let me tell you three is the right answer. Well done. He's the right answer. So first of all summation of our factorial from one to hundred. Let me write down few terms. Of course, I would not write till hundred. Okay. Da da da da till hundred factorial. Okay. Now, when you're dividing by 15. 15 is basically made up of three and five. Correct. Now the moment both these numbers appear in a factorial. From that number onwards, there will be no remainder at all. Right. For example, here one. This is one into two. This is six. This is 24. But the moment you reach five factorial guys remember five factorial is one into two into three into four into five. Okay. That means three and five have started making their appearances. So such numbers will never leave a remainder. So if you go to six factorial, okay. Six factor also three and five has already made its appearance. So till you reach hundred factorial, all these numbers will not give you a remainder. So there will be no remainder at all or you can say remainder here is zero. Reminder here is zero because each of these numbers are divisible by 15. Okay. So this is also divisible by 15. This is also divisible by 15. This is also divisible by 15 and so on. So every number after here will be divisible by 15. So anything divisible by 15 will not give you a remainder. So if a remainder has to come, it has to come from these four guys. Okay. And these four guys add up to give you 33. So the remainder when you get, I'm sure you know this symbol in your computer science. Okay. The remainder that you get is going to be three in the language of congruences. We read this as 33 is three mod 15. That means when you divide 33 when you divide by when you divide 33 by 15, the remainder is three. So your answer is going to be three. Okay. So well done to all of you who got the answer. Very good. Now the major use of the major question which are framed on factorial is a concept which we call as finding the exponent of a prime number finding the exponent of a prime number. Let me call that prime number as P in in anyone. Any question or Krishna also said in n factorial. First of all, what is the meaning of exponent of a prime number in n factorial exponent means guys, we know exponent means power. Okay. So if you want to find out how many times a prime number appears in n factorial, that is what we call as exponent of a prime number in n factorial. I'll give you an example. If I ask you are 24, which is for factorial, right? How many times to appears in 24 or what is the exponent of what is the exponent of two in four factorial? That means if you write this term, which is 24 as a prime factorization, you can see three is the exponent of two in four factorial. That means you can pull out the prime number to three times from 24. Are you getting my point? Everybody's clear with the meaning of exponent of a prime number in n factorial. Yes, sir. Yes. By the way, your notation that I would be using here is E P n factorial. So it means exponent of P in n factorial. Are you getting it? Now, how would I solve this question if this number becomes very large here? Of course, we could do prime factors and we could figure it out. But let's say if I asked you find find e 300 factorial, that means when you prime factorize 100 factorial, what is the power of three that appears in that prime factorization? Correct? How will I solve this? Okay. Now to solve this, there is a small formula which I'm going to discuss with you. Okay. The formula is the exponent of any prime number in factorial of n is given by GIF of n by P plus GIF of n by P square. GIF of n by P cube GIF of n by P to the power four and you continue till you start getting zero. You continue till you start getting zero. For example, let's say 24 itself. If I want to find out what is exponent of two in four factorial, my answer will be GIF please note here. I'm writing down in case you're following this note. You should not get confused. What is square packets? This represents your greatest integer function. Yeah. Now the answer for this will be four by two. See basically this term I have written then four by two square. Okay, if you go to four by two cube, you would realize these terms will start becoming zero after this. Correct. So your answer is only the first two terms, which is two GIF of four by four is one, which is three. As you can see three was our answer. Isn't it? Okay. Now, how do we get to this formula? I'll explain you once I am doing this chapter in detail with you. But as of now just remember it because we don't have much time. We have a lot of things to cover today. Okay. It's very simple. In fact, you can try this derivation and tell me the result personally on my WhatsApp. It's very simple. You just have to pull out the P from the factorial of N. Okay. That is a hint. Let me ask you this question now. What is the exponent of three in hundred factorial? Please give me the answer within one minute. Time starts now. You already know the formula. Oh, shawmik has already given the result. Okay. Aditya also says the same. Vipo also says the same thing. Very good. Guys, 48 is correct. You just have to use this formula. So exponent of three in hundred factorial is just going to be GIF of hundred by three GIF of hundred by three square GIF of hundred by three cube GIF of hundred by three to the power four after this I'll start getting zero zero zero zero. Okay, so don't need to write it because three to the power four is already 81. Okay, so this will become 33. Okay, this will become 11. This will become a three. This will become a one. Okay, and that's it. We don't have to go any further. So 44 plus four answer is 48. Okay, that means if you prime factorize hundred factorial, the maximum number of threes that you can see there in the power of that three number is going to be 48. Let's have another question. Find the number of zeros at the end of hundred factorial. Find the number of zeros at the end of hundred factorial. This is also linked to the same concept 11. No, that's wrong. 11 is wrong. So Shamik Aditya Vibhav Ananya you all are wrong. I think you got tricked over here. I'm sure you would have done this exponent of 10 in hundred factorial, but let me remind you 10 is not a prime number. So the formula will not work. This is very important. I categorically mentioned this P must be a prime number. I understand your understanding your logic here. You want to see how many what is the power of 10 in the factorization of this. But remember 10 is not a prime number. Then how will you do this? So do we find e for a hundred factorial two and five separately and then multiply? Not no need to multiply. See, see what you said was you you are going to find out how many twos is once again somebody is facing a problem. Oh, everybody is able to see the video chats. Yes, sir. Yeah, one second. Prajna is having a problem. I'll suggest to rejoin any issue. Please highlight immediately. Okay. Okay. Yeah, hundred factorial. Let's say have two to the power something. Okay. Okay. Three and all I don't want. I just write five to the power something. Okay. Let's say this power is X and this power is Y. Correct. Now, how would you generate a 10? You would generate a 10 when one two and one five multiply, right? So to generate a 10, you need a two and you need one five, isn't it? Correct. So if I say you have X number of twos and you have Y number of fives, what is the maximum number of tens you can generate? Now, this is a concept which you would have done in chemistry limiting reagent, correct, whichever is lesser in quantity. Absolutely, Ajay, you are correct. Now, whichever is 24, which yes, yes, correct. So whichever is lesser in quantity, that would decide how many tens are getting formed, right? Just like you have it in more concept when two reagents go under chemical combination, basically the compound form depends upon which is in the lesser quantity. Correct. So out of X and Y, which do you think will always be in a lesser quantity? Why will always be less? Why? Because the formula says you have to divide by the number, right? N by P, N by P square and so on. So of course, five will be always giving you a lesser answer. So here, Y will always be less than X. So basically finding the number of zeros, number of zeros, in 100 factorial is as good as, is as good as finding E5 in 100 factorial, getting my point, okay? So E5 in 100 factorial will be what? 100 by 5 GIF, 100 by, sorry, 100 by 5 square, 100 by 5 cube the moment you do, it will start becoming zero. Because zero, zero, zero after that. So the answer is 20 plus 4, 24. That means the number of zeros at the end, I'm not saying in the between and all, on the end of 100 factorial is 24. Is that clear, everyone? Is that clear? Rajna has joined again. Rajna, Rajna. Yeah, yeah, yeah. He has joined. Rajna, if you can hear me, can you see the screen now? Okay, let's take this question. Well, last one, find the exponent of 80 in 180 factorial. Find the exponent of 80 in 180 factorial. 44, Siddhartha says 44. 40, okay, I'm getting different, different response. Everybody is basically saying 44. Okay, let's check. First of all, 80 is not a prime number. 80 is not a prime number. Okay, so what I have to do is I have to first prime factorize 80, which is 2 to the power 4 into 5. Okay, now I have to see how many exponents of 2 are present and how many exponents of 5 are present in 180 factorial. And then I have to see which is the limiting reagent. It's like, it's like saying that to produce an 80, 4 moles of, and I'm, I'm writing it in a language of chemistry, 4 moles of 2 plus 1 mole of, okay, so 1 mole of 80 is produced by 4 moles of 2 and 1 mole of 5. I know it will sound very funny to you, but I wanted to get this thing clarified that we are trying to see the limiting reagent in this case. So first of all, I will try to find out in 180 factorial, how many twos are there or what is the, what is the exponent of 2 in 180 factorial? So let us use the formula. It goes to quite a lot of terms. 180 by 2 to the power 6, 180 by 2 to the power 7 is the last term that I would need. Okay, that's going to be 90 plus 45 plus 22 plus 11 plus 5 plus 2 plus 1, which is going to be 176. Okay, now what is the exponent of 5 in 180 factorial? Let's check it out. So it's GIF of 180 by 5, GIF of 180 by 5 square, GIF of 180 by 5 cube, that's it. We needn't go any further. And that's going to be 36 plus 7 plus 1. Okay, that's going to be 44. Guys, now listen to this. Thankfully, 176 and 44 are also in the ratio of 4 is to 1. Are there also in the ratio of 4 is to 1, correct? So when you say 4 moles of 2 reacts with 1 mole of 5, that means 176 will react with 44, isn't it? Basically, I'm multiplying with 44 both sides. So how many moles of 80 would be found 44 only, correct? So the answer is the exponent of 80 in 180 factorial will also be 44. Are you getting my point? Are you getting my point? Guys, just assume, let's say, hypothetically, let's say, if there were only 162s, then what will happen? Let's say this were 160. Then what would you do? What would your answer be? If this was 160 and this was 44, then what would your answer be? 40, absolutely. Because the limiting reagent is now going to be this guy. So only 45 will react with 162s. Getting the point? Okay, so be careful about solving these kind of questions. Next, we're going to jump to the main concept, which is the fundamental principle of counting. Fundamental principle of counting. Fundamental principles of counting. Okay, now there are normally three principles of counting that we study. One is the fundamental principle of addition. Second is the fundamental principle of multiplication. Okay, this is very vital for your problem solving, very important for your problem solving skill. And third is not a fundamental principle, but that is very important, which we call as the principle of inclusion and exclusion, pi. Principle of inclusion and exclusion. This concept is very vital in some of the topics of fermentation and combination. We'll deal with this later. We'll talk about this later. But these two are very important because this you cannot even solve a single problem in this chapter. Okay, so let us talk about first fundamental principle of addition. What is this principle? This principle is basically based out of logic. You cannot prove it. That's why it is called fundamental. Okay, it is so obvious and logic driven that it's very difficult to derive these formulas. Okay, now in order to explain this principle to you, I'll give you an example. Okay, let's say, let's say there is a cinema theater. Okay, there is a cinema theater and you are within the cinema theater. Okay, now to get out of the cinema theater, there are three exit doors on this side. Okay, and there are four exit doors on this side. Okay, so after the movie is over, let's say you want to exit the cinema theater. So three doors on this side and four doors on this side. Now my simple question to all of you is how many ways can you exit the cinema theater? I'm sure you will not even take a second to answer this. Seven ways. What did you do? You added three and four. You added three and four. Why didn't you multiply three and four? Why didn't you divide three and four? Why didn't you take three to the power of four? This is something which is very fundamental, right? If you have to explain this to somebody, you know, why didn't I multiply or why didn't I take a power? It will look very odd. It's very difficult to convince that guy. You can only say that, hey, I could have taken this door or this door or this door or this door or this door or this door or this door. The moment you are using the word or the moment you are using the word or or you take cases, such words when you come across, you basically add. Are you getting my point? So in general, in general, I can say that if there is a task, okay, let's say there's a task T and this task can be accomplished by doing any one of the sub tasks independently. Any one of these sub tasks independently. Let's say there are n sub tasks. So even if you do T1, your task is done. Example, let's say even let's say T was the task of coming out of the cinema hall and T1 was you taking door number one over here. Let's say this is door number one. Okay. So once you take door number one, your task is over. Isn't it? You are out of the cinema hall. Correct. So if your task could be done by doing any one of these sub tasks independently, so I'll write down where T is done through any one of the eyes independently, independently means independent of each other. Okay. Then we say the number of ways of doing the task T is actually the sum of the number of ways of doing each of these sub tasks. Are you getting it? Let me use the word m over here because n is already being used. Are you getting this point? Okay. Let me give you another example. Let's say in your class, in your class, let's say, in your class, there are five girls and let's say seven boys. You have to choose one class leader. How many ways can you do it? No, sorry, 12. Five plus seven, 12. Why did you add it? Why didn't you multiply it? How would you explain it to somebody? You'll just say by fundamental principle of counting, if the class leader could be chosen from a boy or from a girl, basically, then you add the total number of ways and get your answer. Are you getting it? Yes. Now, what is fundamental principle of multiplication? Let me again take a simple example. Same cinema hall, same cinema theater. Okay. Now this time you are outside the cinema theater waiting for you to get inside to see the movie. Okay. Now there are three entry doors here. You can only enter through these three doors. Okay. E N 1, E N 2, E N 3. And there are four exit doors, four exit doors. E X 1, E X 2, E X 3, E X 4. Okay. So after you can only enter through the entry door and you can only exit through the exit doors. So my question is how many ways can you enter the cinema hall and exit? 12 ways. 12 ways. What did you do here? You multiplied the result. Yes. Because see you would say that I can enter through the exit and exit from this or I can enter from this and exit from this, enter from this and exit, enter from this and exit, enter from this and exit, enter from this and exit and moment you're saying and and and basically you are using fundamental principle of multiplication. So whenever you use or or cases you add them. Whenever you use and you multiply them. So in general, in general, we say that if there is a task T which can be done only when you do all these subtasks T 1, T 2, etc. That means only when all of them are done then only this task will be completed. For example, here my task was to get in and get out. So getting in was my task. Getting out was my sub task to and both have to be done. Right? So in that case, the number of ways of doing the task would be the product of the number of ways of doing these subtasks. Are you getting this point? Let me give you another example. Let's say you want to go from Bangalore Bangalore to Chennai. By a Hyderabad. I know they fall in different directions. Okay. But there are, let's say, he roots from or let's say four roots from Bangalore to Hyderabad. They are four buses. And there are two roots from Hyderabad to Chennai. Correct? How many ways and you have to go through Hyderabad? You cannot go directly from Bangalore to Chennai. In how many ways can you go from Bangalore to Chennai? Eight. Eight ways. Four into two. Now nobody teaches you when to add and when to multiply. It is just that when you read the statement in your mind, it will be clear enough to tell you whether you have to add or you have to multiply. You know, I keep on telling you that you have to add or you have to multiply. It will be clear enough to tell you whether you have to add or you have to multiply. You know, I keep on getting these questions from students later on in class as well. Sir, how do I know how to add or how do I know how to multiply? Guys, just follow your fundamental principles. So you need to go from Bangalore to Hyderabad and the moment you're saying and multiplication will feature in. And you have to go from Hyderabad to Chennai. So four into two. Eight ways to go from Bangalore to Chennai. Are you getting the point? Okay. I would like you to solve some questions first of all. Okay. I just want to see whether your fundamental principle is clear. Hope you all can read this question properly. A five digit number is formed by one, two, three, four, five without repetition. Find the number of numbers thus formed divisible by four. The last one is four and two. There are two ways of writing things. Repetition is there. There is a meaning there. Is it ten? Ten, no. Huh? Somebody's mic is on. Aditya see. Aditya, mute yourself. We can hear your emotions. No, I think Siddhartha's mic is also on. Okay. Many people have started responding. Somebody said 24. Neha is the one who said 24. Siddhartha is also saying 24. Okay. Trippan saying 18. Sushant saying 12. 20. Okay. I'm getting different types of answers. Okay. Time to discuss. Well done guys. Basically I'll what the thing which I like about all of you is you're all making sincere attempts to solve it. Okay. So see here, if you want to make a five digit number which is divisible by 24, what is the divisibility criteria for division by four? A number is said to be divisible by four if the last two digits if the last two digits of the number are divisible by four. Simple as that. Okay. Now out of these numbers, let's say case one. Let's say this is my five digit number. Okay. What are the possibilities I can have? I can have a 12 in the last two digits. Then only it can be divisible by 20. I do it by four. Case two. I can have 24 in the last two digits. Correct. Case three. I can have 32 in the last two digits. Case four. I can have 52 in the last two digits. Okay. Only under these four cases will I be able to form a five digit number without repetition. Look at this. We have to do it without repetition. No repetition. Okay. Now how many ways can you fill these three digits? So now two of them are gone. So three of them are left. So you can fill this in three ways. You can fill this in two ways and you can fill this in one way. What does fundamental principle of counting says? Will I add a 32 and one or will I multiply 32 and one? So your answer will be, of course, we'll multiply them. Why? Because till you fill the first place and the second place and the third place, your job of making that five digit number will not be over. Are you getting my point? So when I'm saying that statement in my mind, I'm automatically using the word and and and every time. So till I fill the first place and I fill the second place and I fill a third place, my job of making that number will not be over. Now I don't have to choose these two because they're already chosen. They're already chosen. They're already chosen. You don't have to choose them. So the number of choices here will be three into two into one, which is nothing but six. In a similar way, do you agree that this will also be three into two into one? This will also be six. This will also be three into two into one. This will also be six. This will also be three into two into one. This will also be six. Okay. Now since these are cases here, these are case one, case two, case three. What will I do with these numbers? Will I add them or multiply them? Add them. Most will add them. So 24 is the right answer. So bad luck to people who have given different answer. Anyways, good try. Let's see whether I can pick up a question for you next. Let's take this five different digits from the set one, two, three, four, five, six, seven are written in a random order are written in a random order. How many numbers can be formed? How many numbers can be formed using five different digits from this set if the number is divisible by nine? Guys, I think they have written a lot of things. They just wanted to form a five digit number where your digits will come from one, two, three, four, five, six, seven. All digits must be different so that the number is divisible by nine. Three sixty. Let me tell you it's not correct. Five eighty. Five eighty is also not correct, Aditya. Oh, somebody says two forty. Two forty is the right answer, Aditya. So I think Aditya has given the right answer but never mind. I'll give you one more minute to just figure this out. Shashank also says two forty. Yeah. Two forty is right. Those who are not getting two forty. Can you just give it a try once more? Okay. See, let's discuss this. Again, what is the divisibility criteria for a number to be divisible by nine? All the numbers should add up to a number that's divisible by nine. Absolutely. Divisibility by nine means some of the numbers, some of the digits must be divisible by nine. Now if you see one plus two plus three plus four plus five plus six plus seven itself. Okay. It adds up to give you three, six, ten, twenty one, twenty eight. Correct. Now you have to drop any two number from this such that this drops to a multiple of nine. Now the next multiple is twenty seven. Right. So for that, if I have to drop the minimum two numbers that will be three so I can never drop to twenty seven. Correct. Okay. Next I can drop two is eighteen. Correct. To drop to eighteen, I have to drop ten. I have to drop a sum of ten. Okay. So I can either drop three seven or we can drop or we can drop six four or four six. Isn't it? These are the only two possibilities. Are you getting my point? Now nine again is not possible because the minimum that you will get is the sum of the first five, which is going to be more than nine. I'm sure. So it's going to be twelve sixteen. Okay. So I cannot get this. The only possibility is I can drop to eight. I can make the sum as eighteen and for that either I drop three and five. Sorry. Three and seven or I drop four and six. So if you drop three and seven, that means you're trying to form a number from one, two, four, five, six. Yes or no. Okay. So this is the case when you drop three and seven. When you drop four and six, you're trying to form a five digit number from one, two, three, five and seven. Yes or no. So there are two cases. Case one. You have to form a five digit number by using one, two, four, five, six. Okay. So you can choose for this place in five ways. This has four. This has three. This has two. And this has one. So basically you need to multiply it. Then only your number will be completed. And by the way, this just becomes five factorial. So see five factorial is a convenient notation that you can use to represent this. Ultimately what you're using, you're using fundamental principle only. But these notations just make your life simple while expressing it. That's why I taught you as a tool just before we started this case to case two is when you want your number to be formed from one, two, three, five and seven. So again, it doesn't matter. It will still be five into four into three into two ways that you can fill up these digits. Okay. So again, it's five factorial. What do I do with these two numbers? Add them because they are cases. So that is nothing but two into five factorial, which is nothing but two into 120, which is 240. All of you must know the factorial till seven at least. So I think till five factorial, I already told you 126 factorial is 720. Seven factorial is 5040. Is that fine? Next question. Let us try what question is there. Dictionary question. Dictionary question. The letters of the word Zenith are written in all possible way. If all these words are written out in a dictionary, by the way, no repetition. Okay. No repetitions of any alphabet. Then find the rank of the word Zenith itself in that dictionary. I'm sure you would have done me type of questions in your school as well. So everybody understood this question, right? See, just like I'll give a simple example. Cat. If I write all possible three alphabet words of cat, what is the first alphabet I'll write? ACT. Okay. Next, I would write ATC. Next, I would write CAT. Next, I will write CTA. Okay. Next, I will write TAC. Then TCA. So out of the six numbers, out of the six, the rank of cat is the third word. It's my third word. Okay. In a similar way, if you write all possible words from the letters of the word Zenith, by the way, how many words will be formed from Zenith? Six factorial. 720. 720. Out of that 720, what is the rank of the word Zenith itself in that dictionary? It's all this. Okay. 615 is what people are saying. 128th. 614 is what Trippan is saying. Sorry. Okay. Let us discuss this. Okay. If you write the very first word from this entire alphabet, the first word will be E-H-I-N-T. Z. And last will be Z. Correct. Now, if you want to find out the rank of the word Zenith, basically you have to see how many words are actually getting formed from the alphabet E-H-I-N-N-T. Okay. So let's start with this. How many words do you think will start with an E? So it's a six letter word, right? So you can choose the other one in five factorial. Now start. 120. That's 120. How many words start with H? Again, you'll say five factorial, 120. So these words will definitely lie above Zenith in the dictionary. How many words starts with I? Again, five factorial. How many starts with N? Again, a five factorial. And how many starts with T? Again, a five factorial. Correct. Now, the moment you reach a Z, we have to be careful because Zenith also starts with Z. I cannot blindly state a statement that all words starting from Z will be above Zenith. No. Then I have to choose next in alphabetical order, which is E. Guys, remember, E-Zenith also starts with Z-E. Correct. So I have to choose the next alphabetical order, which is H. How many words start with Z-E-H? Three factorial? Six. How many words six? Correct. Next is Z-E-I. How many starts with Z-E-I? Again, three factorial, which is six. Correct. Next is Z-E-N. The moment we read Z and N, we have to be careful because Zenith also starts from Z-E-N. Correct. So the next word in line will be what? H. How many words starts with Z-E-N-H? Two factorial, which is nothing but two. Correct. Next will be Z-E-N. I, if I take, Zenith also starts with I. Is it it? Correct. So we have to first take I-T and then take T-H. So one word here, one word here. This is the position of Zenith. So let's count. Sorry. So one twenty into five is six hundred. Six hundred plus twelve plus two plus two. Correct. How much is this? Six hundred and sixteenth word is Zenith. Let's see who gave the right answer for this. Krishna gave, Ranganath gave, Aditya Narayan gave. And most of you were out by a little margin. Okay. Would you like to try one more of this kind? Yes sir. Since Krishna gave a right answer, let me give you a word Krishna. If you arrange the letters, if you arrange the words, if you arrange the Krishna alphabets to make different words from it and place them in a dictionary, what would be the rank of the word Krishna himself? Find the rank, find the rank of Krishna in that dictionary. Be careful while doing it. Krishna without H, right? Yes, yes, yes. 322 is what Sharmic says. Not commenting with this right or wrong. Let's get it out from others. Okay. He changed his answer. 180. That seems too less. 327, 324, 352. Okay. 613. Okay. I'm getting different responses from all of you. So let us first arrange the letters of the word Krishna in alphabetical order. So first will be A, I, J, K, N and S. Okay. So let us first see how many words can be formed from A. Okay. Of course, five factorial words. How many words can be formed from I? Again, five factorial words. They will be definitely above the word Krishna. Now, the moment you put a K, you have to be careful because Krishna also starts with K. So choose the next word in line, which is KA. KA, the number of words would be, the number of words starting with KA would be four factorial. Okay. Next is KI. That will also be four factorial. Next is KN. That will also be four factorial. And next one is KR. You have to be careful because Krishna also begins with KR. So you have to choose the next one, which is A and that's going to be three factorial. Okay. Then KR, I. Krishna also starts with KR, I. KR, I. A, how many words start with two factorial? Okay. Then KR, I. I cannot choose A here. A is already, sorry, KR, I. A is chosen. I cannot be chosen. K cannot be chosen. Then N. Again, two factorial. Right. Then KR, I. S. Now Krishna also starts, but after this we can count it. Next word is AN. That is one. And KR, I. S. N. A. That is one more. So let's count how many words have happened so far. So five factorial, five factorial is 120 into two. Then we have 24 into three. So three into 24. Then we have six. Then we have four plus two. So the answer is 240 plus 72 plus 12. The answer is 320. Fourth word is Krishna. 320. Fourth word is Krishna. Let's see how many of you got this right. And who was the first one to get this right? Aditi Srinoye got it right. Ananya A. Pragana got it right. Siddhartha got it right. And the first one to got it right was Ananya R. Very good. Aditya C also got it right. Shankin also got it right. Wow, well done. Good guys. So with this we are going to start officially with the concept of permutation. Permutations. What is the meaning of permutations? Permutation literally means arrangement. Arrangement. Arrangement of what? Arrangement of certain number of things chosen from certain number of things. So basically we use NPR as a symbol to say that you are arranging our objects chosen from N distinct objects. Guys this word distinct is important here. Please note NPR expression must never be used if your objects involved are identical. Or if some of them are identical. NPR is only used when your objects are all distinct. Okay. Now if you have been given let's say 5 objects. Okay. Let's say I take a smaller example. 3 objects. And I say 3 P2. What do I mean? I mean let's say your 3 objects are ABC. I say choose 2 from there and you arrange them in all possible orders. Arranging means assigning orders to those chosen ones. Correct. So how do you do this? So you consider that there are 2 seats. Okay. First seat you can fill it in 3 ways. Okay. So once let's say one of them is gone. Let's say B is gone. Then the remaining seat can be occupied by anyone of ANC. So 2 ways to fill it. Correct. So what is the total number of ways? Will you add them or will you multiply them? Multiply them. Multiply them because until unless you choose the first place and the second place you are basically not completing your task. So 6 is the number of ways you can arrange 2 objects chosen from 3 objects. Are you getting it? So inherently in the core of the principle what is hidden? Fundamental principle of multiplication is hidden. So all these formulas which you are going to study whether it is NPR or later on NCR they all come from your common sense approach. They all come from your fundamental principle of counting. Are you getting my point? So to get this answer I did nothing but I use my fundamental principle. That's why many a times you can solve these questions without even using these notations at all. Okay. These formulas just come from your fundamental principle of counting. Are you getting my point? Now since we have discussed this expression let us now try to prove that NPR the number of ways to do it is nothing but N factorial by N minus R factorial. Now how do I get this formula? How do I get this formula? Again it doesn't come from the sky. It comes from your fundamental principle of counting. Correct? So let us say that there are N people. There are N people let's say I make a class here. In front of the class I make these R seats. So these are your R chairs. Okay. And this class this is the class by the way. This class has N students. Okay. Now I want to make these R chairs occupied by some students that is R students on this class. How many ways can I do it? Now just start counting from the very first chair. How many ways can I fill the first chair? How many ways can I fill the first chair? The first chair can be filled in any one of the N ways. Basically you can ask any one of the N students to come and occupy that first chair. Yes or no? Now let's say once this is occupied. Okay. This has been taken. How many ways can you fill the second chair? N minus 1. N minus 1 because one of the one of the students is already gone. So remaining is N minus 1. One student can occupy here. How many ways can you fill the third chair? N minus 2. N minus 2. See mark my words. Third chair N minus 2. Fourth chair N minus 3. Who will tell me the R-th chair then? N minus R minus 1. Awesome. Very good. N minus R minus 1. Now according to fundamental principle of counting. Will I add them or will I multiply them? Multiply them. Multiply them because your task of filling all the chairs will not be completed till you make the first chair occupied and the second chair occupied and the third chair occupied and the moment you sing and and and and you are multiplying it. So basically this is going to become N into N minus 1, N minus 2 into N minus R. Let me open the brackets here. Now this is not the way in which we wanted the formula to be. We wanted to convert it to this form. Never mind it is just a simplification. You just step away from the final answer. Now just tell me had I continued with this process what would have been the next term? N minus R. So let's multiply with N minus R and let us divide also with N minus R. Am I doing anything to the expression? No. Whatever I am multiplying with I am dividing it also. Next term would have been N minus R minus 1. So let's multiply and divide with N minus R minus 1. Next term would have been N minus R minus 2. This also N minus R minus 2. Continue all the way till you reach 1. So continue all the way till you reach 1. Now look at the numerator. Don't you see 1 into 2 all the way till N in the numerator? Isn't that N factorial? Look at the denominator. It is 1 into 2 all the way till N minus R. Isn't it N minus R factorial? And therefore this is your formula. Guys again I am repeating this. Please mark my words. This formula is to be used when all the N objects are distinct. Distinct means different. You cannot use this if your objects are identical. Are you getting my point? Okay. We will take a question on this. How many words can be formed? How many words can be formed by using all the letters of the word mathematics? Oh sorry there are some repetitions here. I will just change my question. How many words can you form by using the letters of the word pencil? How many 6 letter words can you make by using the letters of the word pencil? 6 factorial. 720. You would say sir I don't even need the formula to do it. What I will do is I can just choose the number of ways I can fill the first letter here. 6 ways, 5 ways, 4 ways, 3 ways, 2 ways, 1 way. There is nothing but 6 factorial. Isn't it? Some of you would say I would like to use the formula. 6 p6 because from 6 alphabets you are choosing 6 and arranging them. Please note both of them are same because 6 p6 is 6 factorial by 6 minus 6 factorial which is nothing but 6 factorial by 0 factorial which is 6 factorial by 1 which is just 6 factorial. Okay. Are you getting my point? So your formula has come directly from the use of your fundamental principle of counting. Getting my point. Okay. So let us take few questions just on the use of the formula that we have seen today. Let's start with this question. 22 pr plus 1 is 20 pr plus 2 is equal to 11 is to 52. Find the value of r. 7 is what Shomik says. Anybody else? 1 minute. 7. Okay. Take your time. Take your time. No worries. Okay. I think it is 7. Most of you are in the favor of 7. Ananya also thinks, Aditya C also thinks. All right. Most of you are saying 7. Let's discuss it. First of all, 22 pr minus 1 is 22 factorial by 22 minus this is going to be 22. This is going to be 21 minus r factorial. Okay. Divided by let me write 20 factorial and this minus this will be 18 minus r factorial. Okay. This is 11 by 52. Correct. Now, first of all, 22 factorial and 20 factorial. What I can do is I can write 22 factorial as 22 into 21 into 20 factorial. This is 18 minus r factorial. Similarly, 21 minus r I can write it as 21 minus r 20 minus r 19 minus r 18 minus r factorial. Now let's try to cancel few terms which can be cancelled 20 factorial 18 minus r factorial 18 minus r factorial is 11 by 52. I think this also will go by a factor of correct. Now cross multiply. So 21 minus r 20 minus r 19 minus r is 2 into 21 into 52. Okay. Now see here, you don't have to solve a cubic equation. We'll act a little bit smart over here. We'll have to split this term as three consecutive terms and those terms are basically your natural numbers only. Okay. Or so as to say positive integers only. Okay. See the three consecutive terms if you see carefully 19 minus r 20 minus r 21 minus r they're three consecutive terms. So I have to break this as three consecutive terms. So how will I break the C this I'll break it up as three into seven. This I'll break it up as let's say four into 13. Correct. Now let us see what consecutive terms are formed three into four I can see 12 two into seven I can see as 14. So it is nothing but 12 into 13 into 14. Right. Wasn't that a clever move to solve this instead of solving a big cubic equation. Right. Now you compare any one of them for example you compare the least with the least you can get the same answer or this with this you can get the same answer or this with this any one of the three can give you the answer. So let us compare 90 minus r with 12 so r is equal to seven. This is your answer. Is that fine. Is that clear. All right. Next we'll talk about permutations with identical objects permutation with identical objects. Okay. Now there are certain examples of question that would be given to you where you realize that certain number of objects are being repeated. For example if I say how many four letter words can you form from moon. Now your answer cannot be four factorial. This is not your answer. The reason. By two factorial. So 12. Yes. The reason being the two O's are repeated. So if you swap their positions it is not going to generate a new word for you. For example let's say I named it as O1 and O2. So even if you write this as O2 O1 they would be counted as the same word isn't it. Because there is no distinction between the two O's. Correct. So how would I try to tackle those scenarios where we have certain number of objects. Let's say we have N objects. Okay. Out of which P are identical of type one. Q are identical of type two. And let's say R are identical of type three. Okay. So P they are identical objects of the following types. Okay. How do I tackle this. So let's say the total number of ways in which you can arrange these N objects where P objects are identical of type one. Q objects are identical of type two. And R objects are identical of type R is X let's say. Okay. Now I would like you to ask a question. Okay. If let's say these P objects. If let's say these P objects were all distinct. How would this answer change. So first let's say the answer is X. Let's say your answer is X. But now my question is if your P objects become distinct instead of being identical. Then how would this answer change. What would be the answer then. You will say sir simple I would multiply it with P factorial. Yes or no. Yes or no. Because now they can be P factorial arrangement of those P objects within themselves. Correct. No. Correct. Yes sir. If I say in this if my Q objects also become distinct. How would my answer change. You will say further multiplied with Q factorial. If my R objects also become distinct. How would your answer change. You'd say you further multiplied with R factorial. Basically what I've done I have indirectly made all the objects distinct. And if all the objects distinct your answer should actually become n factorial isn't it. In other words you are trying to say X into P factorial Q factorial R factorial is actually n factorial. And therefore your answer becomes. And there's some something what some of you had stated while I asked you that question moon. You divided by the factorial of the number of identical objects in that particular set of objects. Right. So coming back to this your answer for the number of four letter words you can form is going to be four factorial divided by two factorial. Understood. Understood. Yes sir. Now we can take that question mathematics. How many different words can be made by the use of the mathematics alphabets using all the alphabets means taken all at a time. Guys read the question in the exam very carefully. A couple of days back there was a question asked to me if you want to make only four letter words for mathematics how would you make it. That's a different approach. The approach for that is a little bit complicated. Here my question is you are making words by using all the alphabets. This is something different. Okay. So try to solve this. 99. 99. 7. 9. 2. 0. 11 factorial by 8. You just have to give me terms of factorial. Don't simplify it. Yes 11 factorial by 8. 11 factorial by 8 is correct. Okay. So how many words, how many alphabets are there? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. So n is 11. Okay. How many m's are there? 1. 2. 2 m's are there. How many a's are there? Let's ask you. 1, 2. Okay. How many t's are there? Let's say r. 2. 2. 1, 2. Anything else which is repeated? No sir. No. Okay. How many n factorial by p factorial, q factorial, r factorial which is 11 factorial by 2 factorial, 2 factorial, 2 factorial. Okay. No need to actually calculate it. It's fine. Even if you leave it like this. Okay. Next simple question that comes your way is, let's take this. Find the number of permutations of the letters of the word daddy did a deadly deed. Let's ignore the spaces in between. Treat it as if this is a one word. So is it 19 factorial by 9 factorial, 3 factorial, 2 factorial, 3 factorial. Yes sir, 19 factorial by 9, 3 factorial. 19 factorial by 9, 3, 2, 3. Yes. Yes sir. Absolutely correct. So how many d's are there here? They are 9 d's. Correct. Correct. How many a's are there? There are 3 a's. 3 a's. How many y's are there? 2 y's. Y's. How many e's are there? 3 e's. Correct. So total number of word is 19. Total is 19. So since these words are or these objects are repeated, the total number of ways would be 19 factorial by 9 factorial, 3 factorial, 2 factorial, 3 factorial. Yes sir. Okay. Simple. Let's take few questions with restrictions. In the number of 7 letter words that can be formed by using the word success so that the 2 c's are together but no 2 s's are together. Sir, 4. 4. 4 is not the right answer. Yes. 24. 24 is the right answer. Who was that? Sir, Chomick. Absolutely correct. Are we getting also say 24? Yes sir. Okay. How do we do such a problem? So first of all, success. You have 2 c's and 3 s's. Correct. Okay. Now this method is important method. All of you please try to pay attention here. I'm going to use 2 methods which I call as string method and gap method. Okay. Both these methods are going to be used in solving this problem. What is string method? String method is a method where whatever objects or whatever alphabets I want to keep together, I tie them with a string. For example, in this case, I want to keep the 2 c's together. So I tie them with a string. Correct. Because I don't want them to be separated. So I treat them as a special alphabet. Correct. Now if you see you have 1, 2, 3, 4, 5, 6 alphabets left. Correct. Okay. Now gap method is used when you want to separate some alphabets or separate some objects. Okay. So let me call this alphabet as special alphabet X. So what I do, I put the alphabets which I want to separate first like this. Correct. In between, there are gaps. As you can see, there's a gap here, there's a gap here, there's a gap here and there's a gap here. Correct. In these gaps, I have to put my S. Are you getting my point? Okay. Now, first of all, in how many ways can I arrange U, X and E? Six points. Three factorial. Three factorial. Absolutely. Correct. Now out of these four gaps, I have to choose any three gaps to put my S. Guys, here comes a bit of understanding about combination but I'm sure it was an easy thing to count also. Okay. So if you have four of these arrows and you want to take three out of them, correct, there are four ways to do it which you would otherwise call it as four C3 because you have already learned NCR in your class. Okay. But if you don't know that also, it is still fine to solve the problem. Correct. So basically, the four, you can say these four gaps, you can choose any three gaps in four ways, isn't it? Or what do you call as four C3? Right. Even though I've not used the NCR formula up till now, you can actually call it as four C3. That's going to be six into four. That is going to be 24 ways. Is that fine? So given that we have already started talking about combinations. We'll talk about this quickly. Okay. Guys, there are a lot of concepts left in permutation also. We'll come back to it because there is a scarcity of time. I'm dumping to the concept of combination. Combination means selection where you just limit yourself to selection. There's no arrangement happening. That means there's no order assigned to what you have selected. You just select and you stop. Are you getting my point? For example, if you want to select 11 players out of 14. Okay. That is what we call as 11 14 C11. You're just selecting it. Okay. If you're just doing selection, this is what you will say. Okay. But if you are doing the arrangement of those objects, that means you're assigning them a batting order, you would say 14 P11. So that's the difference between selection and arrangement. Right? So selection is just choosing and not assigning any order to it. And arrangement means choosing also and whatever you have chosen, you are ordering it. Are you getting a point? Now, NCR, let me tell you when you're using this formula, N must be distinct. You can't use this formula when your objects are identical. Okay. The next question that would arise in your mind that what is the formula for NCR? I'm sure you would have done this formula in your school. N factorial by R factorial N minus R factorial. Correct? From where do we get this formula? Now, the formula of this is very simple to derive. If you say NPR, what is the meaning of NPR? NPR means selection of, selection of R objects from N objects and whatever you have selected, you arrange those R objects among themselves. Correct? No. From here only I'll get my answer. Basically, I'm using my logic. I'm using my logic to get my answer. From here itself, I will derive the formula of NCR. So NPR is as such, selection of R objects from N objects means NCR and means multiplication. Now, how many ways can you arrange R objects among themselves? You'll say R factorial, isn't it? R objects, if you have to arrange them among themselves, you'll say RPR or you'll say R factorial. Correct? So from here itself, I can get NPR as NPR divided by R factorial, which is nothing but N factorial since NPR is N minus R factorial and there's an extra term R factorial is coming over here. And that's how we get to this formula. That's how we get to this formula. Any questions? So in the interest of time, I'll quickly jump to the properties. Okay, NCR is also sometimes called binomial coefficient. Why it is called binomial coefficient? Because you would realize in the binomial expansions, which you will study in some time in the binomial theorem chapter, these terms appear in front of those variables. That's why they're called binomial coefficients. Anyways, we'll talk about the properties of NCR. First property is very obvious. NCN and NC0, both are equal to 1. Please don't get confused. In case of NPN, NPN was N factorial. NP0 was 1. Are you getting it? Next property, NC, if NCX is equal to NCY. Okay, it means two things. Either your X could be equal to Y or your X plus Y is equal to N. Are you getting my point? Some books say this theorem as NCR is NCR minus NCN minus R. So some books write this property also as this. If your R is greater than N by 2, NCR is also NCN minus R if R is greater than 2. Can somebody prove this by logic? Not by formula. I know by formula you can easily show NCR is NCN minus R. By logic, can somebody tell me why NCR is equal to NCN minus R? You choose N objects then the rest of the objects left. It's the same way to the number of ways to leave them. Absolutely correct. So choosing NR objects from N is as good as choosing those N minus R which you don't want to choose. For example, let's say there are 10 students and I want to choose 4 students to represent the school in some competition. So instead of choosing 4 players or 4 students from those 10, I can instead choose 6 which I don't want them to represent. I don't want these 6 to represent the school. So automatically those 4 left are chosen once. So that's why 10C4 is equal to 10C6. Next property. This is one of the most important properties. This property is NCR minus 1 plus NCR is N plus 1CR. This property is actually called as the Pascal's identity. This is called the Pascal's identity. Now why it is called the Pascal's identity? How many of you have heard of the Pascal's triangle? Anybody who has heard of Pascal's triangle before? Those of you who haven't heard about Pascal's triangle, let me tell you how this triangle works. This triangle always starts and ends with a 1. So let's say 1. The next row will be 1 and 1. And in between it is the sum of these two. So if you add these two, it will become a 2. Again 1 and a 1. Between you add these two, you get a 3. You add these two, you get a 3. Again a 1 and a 1. You add these two, you get a 4. You add these two, you get a 6. You add these two, you get a 4. Now treat the n being the row position. So this is your n equal to 0 row. This is your n equal to 1 row. This is your n equal to 2 row. n equal to 3 row. n equal to 4 row and so on. So each is like a row. And each number is the column position. Let's say this number is ncr is 0 c0. The r here is 0. This is r equal to 0. This is r equal to 1. This is r equal to 0. This is r equal to 0. This is r equal to 1. This is r equal to 2 like that. So treat these individual numbers as r equal to 0, r equal to 1, r equal to 2, r equal to 3 places like that. So if you see this number, take any number. Let's say I take the number 3. 3 is present in the third row and r is 1 over here. So 3 is actually 3c1. Are you getting it? And this 3c1 is obtained from the sum of these two. This is nothing but 2c0 and this is nothing but 2c1. That is what this identity actually says. If you see this identity, it basically says the same thing. Put n as 2 over here. 3c and put r as 1. So 3c1 is nothing but 2c1 plus 2c0. Since this entire property is the founding pillar for your construction of Pascal's identity, Pascal's triangle, that is why it is named as the Pascal's identity. Is it clear how this rule works? Okay. Now I would prove this without using any formula just by the use of logic. All of you please pay attention. So I will prove Pascal's identity just by the use of logic. When you are asked to select r objects from n plus 1 objects. Okay. Let's say these n plus 1 objects contains a special person. Let's say I call them as people. Let's say n people, n plus 1 people. These are special people by the name of A, person A. Okay. And there are rest n people. Okay. This a can be any person. So out of those n plus 1 people, let's say there's a special guy, which is Mr. A. Okay. Now I have to make a team of r from these n plus 1 people. So I have to choose a team of r from this n people. Now there are 2 cases. Case 1. Either Mr. A. Either Mr. A is selected in that r. In the r people whom I am going to choose. Or case 2, Mr. A. Mr. A is not selected in that r. Is there any third possibility guys? No, right? These are the only thing that can happen. Okay. So if Mr. A is selected in that r people, that means how many more I require from how many? I need r minus 1 more people, right? From how many people? These n people. So this can be done in n c r minus 1 way. Yes or no? Yes or no? Are you with me or are you left? Yes. You all are there, right? Yes, sir. Now if Mr. A is not selected, that means I need r people from n people. Correct? That means my task of choosing r people from n plus people can either be done independently by doing this or this. That means cases. So according to fundamental principle, you will add them, right? This or this or means addition. So see without the use of any formula or what, you know, expression, I use just logic to prove it. And I want you to be good at it because sooner or later, your problem solving would be dependent upon not the formulae, but your logic to solve them. Okay, formulae are just like tools. They don't help you to solve a problem in the initial part of it. In the initial part, only logic can break it open. Once you've started, you know, breaking the problem open, then your formula will just help to expedite that process. Is that fine? Okay. Sir, could you repeat case one? Case one, you see, ultimately, what do you have to do? You have to select r people, right, from n plus one. So case one is where your a is selected in that r people. Okay, so let's say a is trippin. I want trippin to definitely represent the school. Correct? So he's definitely selected. So I need r minus one more people to be selected, right? Out of how many n left? Remember, trippin was one of those n plus one people. So that's why n see r minus one. If let's say I said, no, trippin will not be selected. He has some other work. So I have only r, n people and r people to select. Correct? Okay, sir. Yes, sir. Which property was this property number three, right? Let's move on to property number four. Property number four. Let me see how many of you are able to get this. NCN, N plus one CN N plus two CN. And so on till N plus RCN is N plus R plus one CN plus one. Because I'll take 15 more minutes of your time because I wanted to finish this part because this is important for your school school test. So hope all of you can bear with me for 15 more minutes. Of course, KVP guys have to stay a little further. Okay. How do I prove this? The proof for this itself comes from your Pascal's identity. Okay. By the way, how does it come? Very simple. Let me write this term NCN as N plus one CN plus one. Can I say both are the same thing because ultimately NCN and N plus one CN both will give you one. Correct. Now, if you add, if you add these terms, what do you get? What is N plus one CN plus one plus N plus one CN? Now recall, recall Pascal's identity which says N plus two CN plus one. Yeah. NCR, NCR minus one is N plus one CR. So basically your N role is being played by N plus one N plus one. Correct. And our role is being played by N plus one. So your formula will be N plus two CN plus one. Correct. Yes or no. Now this N plus two CN plus one will become what? N plus three CN plus one. Correct. Next term would have been NC three CN. This term would have been become N plus four CN plus one. If you continue doing this, the term which will get added to this will be N plus R CN. So if you add these two terms, ultimately this will give you this as your answer. This identity is called the hockey stick identity. This identity is called the hockey stick identity. Why does it call a hockey stick identity? Basically it comes again from the Pascal's triangle. One, one, one, two, one, one, three, three, one, one, four, six, four, one. As you can see, as you can see the sum of these, follow my marker over here. One, two, three adds up to give you six. Not only that, one, one, one, one, one, four adds up to give you four. One and three adds up to give you four. You see a hockey stick is getting formed every time. The structure is just like a hockey stick that you have. Isn't it? It's a hockey stick. This is actually saying the same thing. See, what is this? This is two C. Sorry, this is one C one. This is one C one. This is what? Two C one. Three C one. And this is what? This is four C two. So it's like saying one C one, two C one, three C one is equal to four C two. Something like NCN, N plus one CN all the way till N plus RCN is equal to N plus R plus one CN plus one. Okay. Treat your N as one over here and treat your RS two. Are you getting my point? Any question with respect to this identity? Next property, property number five, which is very important. NCR by NCR minus one is N minus R plus one by R. Okay. This is very important. This is going to be used in binomial theorem as well. This is going to be useful in binomial theorem as well. You can prove it. You can just expand NCR and NCR minus one and you'll get the answer. Sixth property is NCR. I would request you to remember this result. Next property is NCR is N by R N minus one CR minus one. Okay. You can further break this down. You can further break this down. NCR can be written as N by R N minus one by R minus one into N minus two CR minus two. Are you getting it? You can still further break down N by R N minus R by R minus one N minus two by R minus two into N minus three CR minus three. It can keep on going. Basically, it's a formula where you can express a higher value NCR in terms of a lower value NCR. Okay. Now, I will just prove this part of the formula through logic. Sir, could you go up for a minute? Yeah, sure. Is that fine? I'll prove this formula by logic. Let's prove this by logic. I know my user formula. You can easily do it. How is NCR is equal to N by R N minus one CR minus one. Okay. That means prove that R into NCR is N into N minus one CR minus one. Can anybody prove this by logic? Just by logic. No formula. I'll just prove this by logic. Very simple. Let's say you want to choose R people from N people. Okay. And you want to select a precedent. So let's say my problem statement is I want to choose. I want to choose R people from N people and elect a precedent. Elect a precedent in that, you know, from those R people. So how many ways can you do it? You will say simple. First, I would select those R people. Right? How many ways can you select R people from N people? NCR. No doubt about that. And, and means multiplication or means addition. After you have selected those R people, how many ways can you elect one precedent out of them? You'll say R ways because any one of those R people will be eligible to become a precedent. Correct? Can I say the same activity could be done if you first choose a precedent and then choose the remaining R minus one people. Am I making sense? So you first choose a precedent only. So instead of before selecting any, before selecting those R people, you first selected a precedent only. So that can be done in N ways. Because there are N people in front of you. Any one of them can be chosen as a precedent. Since precedent was also a part of those R people, you require R minus one more people to be chosen. From how many left? From N minus one left. Correct? So that's why this expression and this expression are same. Am I making sense? So first here, the steps here was first you chose R people and second step was you chose a precedent then. Then you made a precedent. Okay. Here the order of the steps were different. Here first you chose the precedent and then you chose R minus one people from N minus one people. Am I making sense here? Okay. Let's take questions. I'll start with this question. I know the images is a little bit blurred, but I'll just tell you what is the expression actually. This is equal. Okay. Okay. I'll let it down separately. NCR is 84. NCR minus one is 36. NCR plus one is 126. Okay. Find N and R. By the way, PNC is also very important topic for KBPY. Yes. Let me help you to solve this because time crunch is there. See, in order to solve this, of course we don't know N, we don't know R. So what I'm going to use, I'm going to fall back on that property. What was that property? NCR by NCR minus one. Do you remember this formula? NCR minus R plus one by R. This is a very important formula, guys. Let me tell you, this is going to be used at so many places. You can't afford to forget it. Okay. So let us take the ratio of this and 36. Okay. So NCR by NCR minus one is nothing but 84 by 36. Okay. If you cancel it by a factor of 12, it's going to be seven by three. And this is nothing but N minus R plus one by R. If you simplify this, it becomes seven R is equal to three N minus three R plus three. Correct. Which is nothing but saying three N is equal to or three N plus three is going to be 10 R. This is your first equation. Okay. Similarly, let us take the ratio of let's take the ratio of this and this. Okay. So NCR plus one by NCR. By the way, if you know this formula, you can also get this formula just by replacing R with R plus one. So here if you just replace your R with R plus one. Okay. Expand it. It will give you N minus R by R plus one. This is given to you as 126 by 84. I think it will go by a factor of 42 here, three by two, which gives you let's cross multiply. Two N minus R is three R plus one, which gives you two N minus three is equal to five R. This is your second equation. Two N minus three is equal to five R. Not two equations, two unknowns. Let's try to solve them. So let us divide one by two. If you divide one by two, you get three N plus three divided by two N minus three is equal to two. Cross multiply. So three N plus three is equal to four and minus six, which means N is going to be nine. Getting my point. Put it in two. So two into nine minus three is equal to five into R. So 15 is equal to five R. So R is equal to three. So RN and both are found here. Is that clear? Okay. I would like to just take two questions before I end it up. Let's take first this question. These type of questions would be included for your semester exams. I'll take this as the last question and then we'll move on. Hope you can read this question at delegation of four students is to be selected from a total of 12 students in how many ways can the delegation be selected. If all students are equally willing. Second, if two particular students have to be included. If two particular students do not wish to be together. In the delegation if two particular students wish to be included together only in the delegation and if two particular students refused to be together. And two particular students wish to be together after this problem will take a five minutes break and then resume with the KVP session. If you're if you're typing in your answer in the chat box, please type it like this one. This is your answer. To this is your answer like that. So first of all, I would like you to tell me what is the answer for the first one. If all students are equally willing to participate. How many ways can you select four students out of 12 for 95 that's correct. So first one, the answer would be 495 first one answer will be 12 C4 that is 495 by the way 12 C4 is 12 factorial by four factorial eight factorial. 12 factorial you can write it as 12 into 11 into 10 into 9. Eight factorial eight factorial will get cancelled. Four factorial is 24. 24 can be cancelled by this five so it's 45 into 11 which is 495. Second part. Who's ready with the second part? Sir is it 90? No 90 is not correct. You know why you got 90 Siddharth? Sir 45 times 2. Why did you do 2? Because two of them have to be there and then only you can do. One second. Are you selecting those two students? No, it is already given there are two particular students. Particular means you don't have to select. You don't have to select them my dear. I'm getting my point. Particular means they are known which two are not participating so you don't have to select them. So if you're doing you know into two then it would be wrong. Yes, what is the answer? 10 C2. That's obvious 10 C2. So if two students are included you need the other two from the remaining 10. So 10 C2 done. Answer is 45. Getting the point. Next. Two particular students do not wish to be together in the delegation. 450. Let me write it over here. See how do you do this? Total selection minus where they wish to be together. Where two students wish to be together. If you do that you will automatically account for the case where these two do not want to be together in the delegation. That means either both of them will be absent in the delegation. Or only one will be present. Both cannot be present in the delegation. So the answer is 495 minus 45 which is 450. Absolutely correct. Ananya, Shamik give the right answer. Next. Fourth part of the question. If two particular students again particular you don't have to choose them. Two particular students wish to be included together only in the delegation. So 255. Sorry. 255 is absolutely correct. Awesome. Very good Shamik. Very good. That was very fast. See two students wish to be together only in the delegation means either they both will be present. Or they both will remain absent. It cannot happen that exactly one of them is there in the delegation. So what is the way in which both are absent from the delegation. If both are absent. Remember you have to select. You have to select how many people. If both are not selected how many have to select. 4 from 10 right? Yes. I don't know. Or if both are present. Then it will be 2 from 10. Correct. So 10C4 is going to be how much 10C4 is going to be 210. You can calculate it. 10C2 is 45. So altogether the answer is 255. Excellent. That was very fast. Fifth one. If two particular students refused to be together and two other particular students wish to be together. This is slightly thought provoking. Take your time. No need to rush. Who is playing with that annotation? Yes you can annotate on the screen. So I can see somebody scribbling on the screen. Okay. Since it is taking time. 285. 285 is not the right answer as per the book. Okay. Let's say AB wants to be together. And C and D don't want to be together. So these guys want to be together. Okay. And these two do not want to be together. So what are the possibilities? Either AB is included. And C is there. AB is included. And just D is there. Correct. And just AB is there. Both C and D are not there. Both C and D are not there. Or both A and B are not there. C is there. Both A and B are not there. D is there. And all of them are not there. I think these are the only cases. Am I missing out on any case? Let me name them. So can I do this problem by saying AB is there. And at least, can I say AB is there. And you want exactly one of C and D to be present. That is total case where both C and D are present. Both C and D present. Yes or no? And second is the case where AB is absent. And again total case where both C and D are present. Can I do the problem in this way also? I think this would be a shorter way to do it. So let's say AB is present. AB is present. Correct. So what are the total ways in which AB is present? If AB is there in the team, that means you need to select two more people from the team. Yes or no? Is it fine guys? Now let's say AB C D is also present. So this is AB C D is also selected. So that case will be actually one because you have already selected four people. So let me do it like this. AB is selected. Always selected. The number of ways will be 10 C 2. That's 45. Okay. AB is always excluded. That would be 10 C 4. That is nothing but 210. Okay. AB C D is always selected. That is going to be just one way. Correct. Because because you just wanted four people and all of them is AB C D. Next is AB is excluded and C D is included. This would be 8 C 2 ways. Isn't it? AB is always excluded and C D is always included. That means out of 12 people two people you have already selected. Okay. So you have left with 10. You have selected like this. So C and D is already been selected and you don't want to select A and B. So you just have eight people. Now since two are selected and two more are required, that will be AC 2. Okay. Let me call this as S 1, S 2, S 3, S 4. So can I say the number of ways in which you can select A and B together is going to be the first expression S 1. Okay. Okay. Plus S plus S 2 minus S 3 minus S 4. Right. Why minus S 3? Because they cannot be selected together. C and D cannot be selected together. In both the cases S 2 and S 4. Sorry, S 3 and S 4. So the answer will be 45 plus 210 minus 1 minus, what is 8 C 2? 8 C 2 is 8 factorial by 6 factorial 2 factorial. Which is nothing but 7 into 8 by 2 which is going to be 28. So your answer is going to be 255 minus 29 which is 260. If I'm not wrong. How much is it? 29. 226. Is that fine? This will be your answer. Okay guys. Those who want to drop off the call, you can drop off and just continue with the KBPY course and all the sessions will be recorded. Don't worry and we'll take a 5 minute break. Thank you sir. Thank you sir. Bye sir. Thank you Sankin. Bye. Thank you. We'll just take a 5 minute break. Alright. So people who have sat back for KBPY today I'm going to begin with a coordinate geometry concept which is pair of straight lines. I think straight lines was already done with you in the school syllabus as well as for GE main syllabus. So in KBPY we have an additional concept which is the concept related to straight lines called the pair of straight lines. Okay. So let's talk about it. This is the concept where we are trying to study the nature of the combined equation of two lines. So first of all what is a pair of straight line? Now let's say I have two equations of line given to me. One is a1x, b1y plus c1 equal to 0 a2x b2y plus c2 equal to 0 Okay. I ask you let's say these are the two lines guys let me tell you they may intersect they may not intersect also. So it's not necessary that a pair of straight lines is only formed when the two lines are intersecting. So even to non-intersecting lines if you combine the equation in this manner of these two lines this combined equation is called the equation of the pair of straight lines. This equation would be the equation of the pair of straight lines. Let me tell you there is a particular way in which you multiply it. Before multiplying you should have converted both these equations to their general form. What is the general form? When you write the equation of a line as ax plus by plus c equal to 0 that is called the general form. Okay. I'll give you an example. Let us say I give you the equation like this. x equal to minus y and x is equal to y. So let's say these are the two lines L1 and L2. I say give me the equation of the pair of straight lines. Now if you want to give me the equation of a pair of straight lines first you should bring both of them to their general form. Right now they are not in the general form. So the general form for this would be x plus y is equal to 0. That means you should have a 0 on the right-hand side. All the expressions must come to the left-hand side other than 0. Now when you multiply it you get x squared minus y squared equal to 0. So this equation is the equation of the pair of straight lines these two. Now many people ask me how does it make a difference if you multiply it right here itself. See it will be different. As you can see if you multiply this will become x squared this will become minus y squared. And if you bring this minus y squared to the left-hand side it becomes x squared plus y squared equal to 0 which is different from this case. By the way this is a point which is 0 comma 0 and it is not a straight line. Because two perfect squares cannot add up to give you 0 unless and till both are 0. Are you getting this point? Okay. Now something very interesting here you just now figured out that this gives you a second-degree equation. Isn't it? Now something very interesting that I want all of you to listen because this is going to be useful for other conic section part as well. Normally all these geometrical shapes like circle parabola ellipse I know we have not done this chapter with you but just a quick heads up on these geometrical figures hyperbola pair of straight lines the equations of all these conics or all these geometrical shapes are second-degree equations. Let me tell you. Okay. That means the equation for circle parabola ellipse hyperbola and pair of straight lines looks like a second-degree equation of this nature. A x square plus b y square plus 2 h x y plus 2 g x plus 2 f y plus c equal to 0. Who has made this star over here? Okay. This is called a general second-degree equation. This is called a general second-degree equation. Let me tell you all the conics have the same general second-degree equation. That means they vary only in terms of their a b h g f n c relationship with each other. This overall broadly speaking the structure the skeleton of all these conic section is a second-degree equation. So even a circle would have terms like this. Parabola will have hyperbola will have. Just now we saw a pair of straight lines also has. So a natural question will arise in our mind that if somebody gives me a second-degree equation okay if somebody gives me a second-degree equation like this. 6 x square let's say minus 4 y square plus 5 x y plus 7 x plus 13 y minus 3 and they say and they ask you this question which conic does it represent? Does it represent a circle or a parabola or an ellipse or a hyperbola or a pair of straight lines How will I want? How will I come to know about that? For that I am going to discuss with you a small theory. I will not give you any proof for it because proof anyways I am going to do in the normal class so there will be just repetition of the same concept. So I will come back to this page and go to the next page. The general theory is if you have any second-degree equation given to you like this okay the first thing that we find is we first find an expression which I call as delta delta is nothing but a b c plus 2 f g h minus a f square minus b g square minus c a square all of you please note down this expression this expression I would call as delta note it down Sir why do you write 2 h, 2 c, 2 f, why not just h there is a actually convention that we write it like this okay I will tell you when I derive this why we actually write it like this I understand your concern why we unnecessarily put a 2 in front I will tell you why we write it like this actually I will talk about a circle first you will be able to relate to it much better so if you see a circle let's say I ask you to find the equation of this circle whose center is 1,2 and whose radius is let's say let me not write 1,2 let me write it as alpha,beta then you will understand and let's say whose radius is r okay so if you write the equation of a circle basically you will say circle is nothing but locus of such a point h,k which moves in such a way around this point o such that this distance is always r so you will use locus equation locus equation is x-alpha whole square a-beta whole square is equal to r square correct if you expand it you end up getting x square alpha square-2 alpha h k square beta square-2 beta k is equal to r square okay then you generalize this generalize this by replacing your h with x remember your locus how do you solve locus questions so first you assume a specific point h,k and then you generalize it with h with x and k with y so it ends up becoming x square y square minus 2 alpha x minus 2 beta y correct plus k square, sorry alpha square plus beta square minus r square as you can see there is a presence of a 2 over here along with this there is no xy term though but there is a presence of 2 with x and y that's why we make it a norm of putting 2 here okay this delta is the first thing that you will calculate a b c plus 2 f g h minus a f square minus b g square minus c s square of course a b c h g f c value would be given to you in the question okay now listen to this conditions if your delta is 0 then it implies that the equation represents a pair of straight lines and I am going to tell this without a proof because proof anyways I am going to do with you later on but by the time I do this the q i would have been very k b p y exam would be very close it will be around the corners if delta is not 0 okay and a is equal to b and h term is 0 then this equation will represent a circle that in a circle equation there will be no xy term just now we saw there is no xy term over here there is no xy term correct and the coefficient of x square and y square would be same correct I will give you the proof later on don't worry about it if delta is not equal to 0 if delta is if delta is not equal to 0 and h square is equal to a b then remember the equation would represent a parabola okay again if delta is not equal to 0 and h square is less than a b then this equation would represent an ellipse and if delta is not equal to 0 and h square is greater than a b then this would represent a hyperbola these conditions must be understood properly keep it in mind getting it let us go back to the question hope you have noted these conditions now let's go back to the previous question yeah now tell me what does this equation actually represent remember here your a is 6 b is minus 4 2h is 5 2g is 7 2f is 13 c is minus 3 so let us find delta by the way I have written it in small letters hope you are not confused a is minus 4 hope you are not confused a b c plus 2f g h minus a f square minus b g square minus c h square just check this was the condition which I gave you for delta yeah this condition a b c plus 2f g h minus a f square minus b g square minus c h square let me go back to the previous slide I think the question is in the previous slide okay let's calculate here so what is a b c a b c is going to be 72 what is 2f g h 2f into g h minus a f square minus a f square minus b g square so plus 4 b g square minus c h square so plus 3 h h is going to be 5 by 2 square let's calculate this in order to save time I will just use my calculator but please you don't use it because calculators anyway is not allowed so 72 plus 72 plus 2f g h will be 13 into 35 by 2 minus 6 into 6.5 square plus 4 into 3.5 square plus 3 into 2.5 into 2.5 what are you getting guys I'll solve it again no no no a b has not to do anything with the eccentricity it is related to it but it is related to it I'll talk about it when I do it later on 72 plus 2f g h into 6.5 into 3.5 into 2.5 minus 6 into 6.5 into 6.5 plus 4 into 3.5 into 3.5 plus 3 into 2.5 into 2.5 oh I am getting 0 I am getting 0 as the answer correct what does this mean it means it represents a pair of straight lines so this second degree equation is representative of a pair of straight lines are you getting my point now next question that I would like to ask you if this represents a pair of straight lines tell me what are the two lines which make this pair that means break this up into see basically my question is break this equation into product of two lines so what are these two lines I would like you to find out can anybody solve that see basically doing the reverse activity you are basically trying to see which two lines multiplied will give you this second degree equation how do you do this guys for this there is an approach all of you please listen to this approach in future also you will get questions where in j e also you will get a question where you will be given a pair of straight lines and they will ask you which two lines make this pair of straight lines for that the approach is first focus on the pure second degree terms so please note 6 x square minus 4 y square and 5 x y they are all second degree terms these two are first degree terms these are second degree terms these are first degree terms and this is just a zero degree term okay zero degree term just second degree term try to factorize it okay I am sure you can split the middle term so you can treat it as if it is 6 x square plus 5 x y minus 4 y square right you can take a 24 can be written as 8 into 3 so you can break it up as 8 x y minus 3 x y minus 4 y square so basically what I am doing is I am factorizing it okay so take a 2 x common here you will get a 4 x sorry 3 x 3 x plus 4 y take a minus y common here you will get a 3 x plus 4 y okay so this is factorized as 2 x minus y 3 x plus 4 y okay I am purposely leaving a gap after these terms because it is not only the second degree terms which make up this equation they are also first degree terms and constant now remember those first degree terms and this constant comes because of some constant over here let me call it as L and M if I somehow find this L and M I will be able to find my both the lines which make this equation correct yes or no and how do I get that simple just multiply this back but when you are multiplying be smart don't multiply everything only multiply those terms which would lead to the cause which will lead to the first degree terms okay for example this 7 x how would x come from here x will come when your L multiplies with 3 x and your M multiplies with 2 x right these are the only 2 terms which will contribute to an x right so can I say 3 L 3 L plus 2 M is equal to 7 getting my point how will you get a y term your y term will be obtained when your minus y multiplies with M or your 4 y multiplies with L so that is 4 L minus M and that is going to be 13 can you solve for L M N by these 2 equations of course we can let's multiply the second equation by a 2 okay so this will become I am making a change here itself this will become 26 this will become 2 and this will become 8 okay just add it just add it I am not missed out on any term just check 4 L minus M is going to be 13 yeah I have only multiplied this with 2 so I don't have to multiply again yeah so 11 L is going to be 33 so L is going to be 3 once you get L you can actually get M by the fact that L M is going to be minus 3 as you can see constant term will be obtained when you multiply L with M 3 into M is minus 3 so M is going to be minus 1 so finally your lines that would make this line I am writing it over here will be nothing but 2x minus y plus 3 and 3x plus 4 y minus 1 is this fine guys okay since we only had a half an hour session today I am not I don't want to cover too many things but there are a lot of things left in this chapter which I am going to take in the next online session with you okay by the way before we close for today all the best remember 2 things practice from your NCIT exemplar and in the exam do not make any kind of silly mistake okay and all the best do your 100 percent don't come back saying that I could have done better that's a worst feeling then you know if you don't know a concept that's fine if you know the concept and you do bad then that gives you a worse feeling okay over and out from my side thank you and have a good day write 2 mock tests today and tomorrow thank you sir thank you thank you so much