 distribution of an identical objects into our different groups into our distinct groups or different groups is the the wordings of these concepts they are so similar that you know people get confused sir the first one was distribution of and distinct objects into our distinct groups second was arrangement of and distinct objects into our different groups now we are doing this thing distribution of an identical objects into our different groups okay so get this minor change in the wordings this can make a difference to you solving or not solving the question it's very important okay alright so now let me begin with a question again I don't want to directly give you the formula I am not that kind of a teacher who will do spoon feeding because that kind of learning is short-lived let me give this as a question and let us say there are six laddus okay laddus Diwali season went by I hope you had laddus okay now these laddus are all identical okay assume that they have been manufactured by a machine all they look same okay just like your Ferrero Rocher chocolates they will all look the same okay so let's say I'm just doing an alienization of the chocolate let's say we have laddus okay and these laddus have to be distributed between three needy children let's say four children they want to have laddus okay this means this is a girl child so you are distributing these laddus to these three children how many ways can you do that so basically it's a case where you are distributing seven sorry six identical laddus into three distinct people how many ways can you do that think first now there's no restriction so let's say blank group is allowed blank group is allowed and I'll take one more case blank group is not allowed blank group is not allowed think carefully and then answer how will you do it hint is take a clue from the previous discussion that we had six that will be very very you know useful in this in this regard so you have six identical laddus we have to split this into three groups blank group is allowed how will you do it mother think more deeply see guys if you have to if you have to split this into three groups all you need to do is use two sticks correct wherever you place the stick that will automatically become a division of the chocolates so let's say I drop a stick here and drop a stick over here what does it mean it means the first child will get one laddus second child will get all the four laddus here in between and the third child will get one more laddus correct so all you need to do was try to arrange try to arrange six identical objects of type one with two sticks you can call it as a type two object so eight objects are there out of which six are identical of type one and two are identical of type two so what will you how many ways you can arrange it your answer would be clearly eight factorial by six factorial two factorial right so basically you'll say eight C2 correct other words you can also say sir I can put the sticks in these places correct yes or no but this will not help you to solve the question because what if I put the both sticks in front or what if I put both six in the back are you getting my point are you getting my point okay or I could put multiple sticks in between so this would not be a right way to solve the question probably this will be helpful in the second case where blank groups is not allowed okay so the number of ways here would be just eight factorial by six factorial two factorial which you can call it as eight C2 or eight C6 both are acceptable both are acceptable okay now if you extend this to n objects and you are distributing those n objects to our people the formula becomes n plus r minus 1 c r minus 1 or you can say n plus r minus 1 c n both will give you the same result this formula is to be noted down and memorize this is the only formula which I'm asking you to memorize of all the things we have discussed so far because this is heavily used heavily used formula very heavily used okay at least 10 times more used as compared to the other formula that we have studied so far okay is this fine correct okay now with the blank group is not allowed then you can say you have to place the sticks in these gaps correct so you have to choose you have to choose r minus 1 gaps from n minus 1 gaps where you will place those sticks the moment you place those sticks automatically groups will be formed such that no group is blank so this is another formula however I still say this formula is the one which you need to remember others formula you may tend to forget also it will not do any harm to you is the idea clear okay let's take questions questions are very very important different questions we will take under this so that the concept is cleared okay so let's start with this question four boys picked up 30 mangoes four boys picked up 30 mangoes in how many ways can they divide them if all mangoes are identical now since no information is given you can consider that blank groups is allowed that means there may be a boy who is manguless sir which word is manguless sir who is without a mango okay simple okay so let's say boy 1 gets x1 boy 2 gets x2 boy 3 gets x3 boy 4 gets x4 total should be 30 okay and it is like your r is 4 n is 30 and blank groups is allowed so I can directly use this formulas that's the reason why I asked you to remember this formula so it'll be 30 plus 4 minus 1 C 4 minus 1 that will give you 33 C 3 we can also write it as 33 C 30 both of it okay now the reason I wrote it like this is because let's say if I frame a question like this can I go to the next slide can I go to the next slide okay let's see if I frame now let me let me use the same slide let's say if I frame a question like this find all non-negative integers x1 x2 x3 x4 such that x1 x2 x3 plus x4 is 30 please let me tell you let me enlighten you here this is same as this question so in J they may change the format of the question but the underlying concept would be the same so finding all non-negative integer solutions for this equation is as good as asking how many ways can you distribute 30 identical mangoes to four people are you getting this point are you getting this point so the number of ways you can distribute these each of them will form a solution to this correct so let's say you distributed it like 0 0 15 15 or you distributed it like 15 15 0 0 so each one of them will form us different type of solution for this equation okay so be aware whenever a question related to this is asked don't get confused the underlying concept is the distribution of identical objects into different groups got the point let me give you let me bombard you in fact with more such questions alright let's solve this question everybody alright find the positive number of solutions of this equation under the following condition zero values of XYZ are included if 0 is included it is like blank groups are allowed okay so the first case the answer would be simply n plus r minus 1 c r minus 1 where n value is 20 our value is 4 so the first answer will be 20 plus 4 minus 1 c 4 minus 1 which is 23 c 3 okay yeah no need to calculate the values I'm fine with the expressions okay second is 0 is excluded now this is something which can be directly related to the concept that blank groups are not allowed if blank groups are not allowed now all of you see there are two ways to do it if blank groups are not allowed basically there are 20 objects okay there are 20 objects and in between them there are 19 gaps correct you have to place 3 6 in these gaps so that will be 19 c 3 correct so this will be your answer this is one way to do it other way to do it is you can say you can replace your x with some other variable plus one you can replace your y with some other variable plus one z with some other variable plus one and w with some other variable capital W plus one correct and rewrite the same equation with the new variables okay so this will become x plus y plus z plus w equal to 16 now many people ask me why did you do this because when you say x is greater than equal to 1 it automatically means capital X is greater than equal to 0 so this would be a case where you are trying to distribute the objects where blank groups are allowed so again you can use the formula n plus r minus 1 c r minus 1 and here would be 16 r would be 4 and this will become 19 c 3 same answer I got in a different way are you getting my point here okay now why I told you this approach is because your questions may be slightly more complicated let me take a version which is slightly more complicated let's say I take this question here you can see that each variable has its own constraint okay so they're asking you for the integral solutions but x should be greater than equal to 1 y should be greater than equal to y should be greater than 1 by the way y greater than 1 is like y greater than equal to 2 because integral solution is what they're asking correct okay so each variable have different different constraints on them are you getting my point okay so the way to solve this would be I mean let me take this as an example for you because I know most of you can solve it on your own so what I can do is I can replace my x with capital X plus 1 I can replace my y with capital Y plus 2 I can replace my z with capital Z plus 3 and let TB as it is so we can say that saying this is as good as saying this and we can say that saying this is as good as saying this and saying this is as good as saying this okay so and of course T is anyways greater than equal to 0 okay so you can change the entire equation to be capital X plus 1 capital Y plus 2 capital Z plus 3 plus T equal to 29 most of you would say sir this can also be seen as if you are giving one object to X 2 object to Y 3 object to Z in the beginning itself yes of course I'm doing the same thing but I'm just writing it in a more mathematical fashion so you can think as if you are trying to solve this equation I think it's 23 I believe right okay so now you can use this as your n value okay r is already 4 so your answer will be n plus r minus 1 cr minus 1 that's nothing but 23 plus 4 minus 1 C 4 minus 1 which is 23 sorry sorry 26 C 3 is this clear so this approach really helps you a lot when you're trying to solve questions like this okay now there are few different versions of this concept let me give you another version first please note this down please note this down done noted alright let me ask you this question find the non-negative integral solutions to the equation X plus Y plus Z less than equal to 29 how will you solve this question now I'm sure most of you would try to make 29 cases out of it that is when X plus Y plus Z is 0 when X plus Y plus Z is equal to 1 da da da da till X plus Y plus Z is equal to 29 and you would like to find all the solutions for these and add them of course that is one of the method but there's a smarter way to do these kind of questions so whenever there's a less than inequality involved we use an approach called the dummy variable yeah dummy group absolutely we use the approach called the dummy group okay so what I'm going to do is I'm going to write the same thing with a dummy variable and equate it to 29 now what is this dummy group this dummy group is a value which is greater than equal to 0 right so let's say if I have to get a 29 I can have it like and I'm just taking a small example so let's say this is 10 this is 6 this is 2 and let's say this is one right so anyways if you just see X plus Y plus Z it will be 28 only 28 means it will be lesser than equal to 29 so whatever value you assign to this dummy group if it is greater than equal to 0 automatically the other three will add up to a value less than 29 if it is equal to 0 it will become equal to 29 so what I'm trying to say is by changing the value of D as 0 1 2 all the way till 29 you are getting all these equations correct so I can treat this equation as if I'm trying to solve this so n is 29 R is 4 and your answer will be 29 plus R minus 1 C R minus 1 which is 32 C 2 sorry 32 C 3 is this understood any question on to how to solve this so let me take another condition another condition let's take another question question is let's say you are rolling three dice you are rolling three dice okay find the number of ways find the number of ways in which you can get a sum of 9 on the dice in which you can get a sum of greater than 9 sorry some off in which you can in which a sum of greater than 9 can be obtained on the die how will you solve this question think think about it and let me know yes Aditya so now I'm trying to solve the case where x plus y plus z is greater than 9 so let's say this is the number which appears on the first day okay there's a number of okay let me do one thing okay fine let's let's take this question so this is the number which appears on the first day this is the number of number which appears on the second day and this is the number which appears on the third day how will you solve this question this question looks very simple but it will give you a lot of learning so let me solve this question for you now here the problem is here the problem is first of all you're solving an inequality okay second problem is each of these values actually lies between 0 to 6 each one of these values will actually lie between 0 to 6 correct so even if you say I take total cases and I will subtract those cases where x plus y plus z is less than equal to 9 then let me tell you even finding this will be a challenge because because there is an upper restriction on these days are you getting my point right are you getting this this is very very important because if I say less than equal to 9 there would be a case like 8 0 0 right oh sorry it can't be 0 also very good yes yes sorry sorry Aritra yeah it has to be between 1 to this thing so I cannot have a case like 7 1 1 right because that would come under this right so this is also not going to help me to solve the question so now let me discuss with you something very very important here which is called multinomial theorem multinomial theorem okay all of you please pay attention to this because this is slightly on a complicated site multinomial theorem is useful when there is an upper restriction on the variables upper cap or upper value restrictions on the variable lower value that is not a problem lower value you know how to you know tackle it we just now did know x was greater than 1 y was greater than 2 so those were lower value restrictions on those variables if there's an upper value restrictions how it is useful so see what I'm going to do I'm just going to take this as an example to solve it so let's say I'm trying to solve this in the you know example where there is a restriction of values on x and y and z let me write it together I don't want to waste time writing it again and again okay with this restriction okay so first of all the normal thing that we were doing that we'll do we'll put a D over here and equate to 9 now D only has a restriction that D should be greater than equal to 0 right so there is no other restriction on D its upper limit can be anything lower limit can be greater than equal to 0 okay now all of you please pay attention see I'm going to construct a multinomial term Gayatri please do not do this Gayatri no writing on the screen please it's hanging so are you writing on the move are you seeing this video on the mobile phone laptop so unless until you decide to annotate with your pen how will that sign come okay I can understand if you're doing it with a hand on the mobile okay please do not annotate on the screen okay that deserves everybody okay now everybody please pay attention now this x can take values from 1 2 3 4 5 6 correct this y can take values from 1 2 3 4 5 6 this z can take values from 1 2 3 1 2 3 4 5 6 correct and this D can take any value from 0 1 2 3 of course it can go to 9 okay now all of you please pay attention these numbers which I have written I'll be putting a variable underneath them let's say I'd put a T okay this numbers also I'll put a variable underneath them now why I'm putting them I'll explain to you in some time just first follow what I'm trying to do okay and I group them that means I'm multiplying these okay next term also okay and the next term also so basically these numbers which these variables can take I'm raising it every number to a power of T now why am I doing it all of you please listen to this see when you're multiplying these terms every term here will multiply with every term correct so let's say when you're multiplying it you will get all the different powers of T I think the minimum will be T to the power 3 correct okay and maximum will be T to the power let's say 6 6 6 9 which is 27 right isn't it minimum that you will get will be T to the power 3 correct when the very first terms of each of these bracketed terms will multiply correct now these terms will occur with a coefficient okay coefficient mean there will be some number in front of them right similarly there would be a term like 9 also T to the power 9 with some number in front of it now why I chose T to the power 9 I'll explain you in some time first listen this out now these numbers that you see on the top here is basically the sum that you can obtain on or the sum of the numbers that you can obtain from adding x y z and D for example if I take 1110 that's the minimum I can get that's the minimum I can get and that would be your T to the power 3 coefficient so the number of ways in which I can get a 3 this coefficient keeps a count of it so it gives it keeps a count of the number of times you will obtain T to the power 3 term that is to say the number of times your x plus y plus z will become equal to or you can say x plus y plus z plus d will become become equal to 3 are you getting my point here okay so each of the terms that you will get when you multiply they will have some coefficient in front of it right see I'll give you a very simple example to relate to it when you do T to the power 0 T to the power 1 again when you multiply with T to the power 0 T to the power 1 what do you get you get T to the power 0 you get 2 T to the power 1 you get again T to the power 2 there are numbers in front of these correct 1 2 1 these numbers will tell you in how many ways T to the power 0 is produced when you multiply it 2 means T to the power 1 is produced 2 times are you getting my point here so these numbers which are your coefficients they basically keep a track off or keep a count of how many ways that particular power of T is being generated got the point so why am I keeping a count of it because if I know how many ways T to the power 9 is generated that means if I know this number my job will be done correct let's say I pick up a 3 from here I pick up a T to the power 2 from here and I pick up a T to the power 1 from here and I pick up a T to the power again a 3 from here can I say that would be one way to generate T to the power 9 and the numbers that you see as a power of T here those will be your x y z and d values are you getting this fact so I'm trying to indirectly link it to multinomial theorem concept which is something which you are going to study after binomial theorem so there's a small half an hour session I'll take on that multinomial because there's so many terms involved okay and this theorem basically is something which we are going to discuss with you when I'm doing the binomial theorem chapter but as of now that theorem is not required but I'm taking the help of it to solve this present scenario okay so is this understood why I am looking out for the coefficient of T to the power 9 in this expansion because if I do that I will get to know how many times T to the power 9 is being generated indirectly I'll come to know in how many ways x plus y plus z plus w will become a 9 that is what we are looking for over here isn't it correct now having said so is it that easy to find out T to the power 9 over here because you have to multiply so many terms no you don't have to multiply so many terms it is really easy to find T to the power 9 from here I'll show you how so let me go to the next slide I'll carry the same problem in the next slide so my job is to find the coefficient of T to the power 9 in for the purpose of saving time I will just write this to the power 3 and this to the power till 9 okay now all of you piece by engine this is actually a geometric series okay both are geometric series both are geometric series yeah sorry for the slip of pen so this is a geometric series and we all know in geometric series most of you would have done this in school also that if you have a a r square a r a r square all the way till a r to the power n minus 1 the sum is going to be a 1 minus r to the power n by 1 minus r correct so this we can write it as T 1 minus T to the power 6 because there are six terms okay divided by 1 minus T whole to the power of 3 this also I can write it as 1 minus 1 1 minus T to the power 10 by 1 minus T like this okay indirectly I'm finding coefficient of T to the power 9 in this expansion now what is this expansion I'll discuss with you step by step so indirectly we are finding coefficient of T to the power 9 in T cube 1 minus T to the power 6 whole cube 1 minus T to the power 10 and 1 minus T to the power minus 4 okay so what I've done please pay attention I've just taken the cube of this term so that will be T cube 1 minus 6 to the power cube denominator term I have left out as of now this is again 1 minus T to the power 10 and the denominator terms combined I have written over here does it make sense this is a complicated question actually but one example is sufficient for you to understand the whole process don't worry about it okay now let us make our life simpler if I'm finding T to the power 9 in this and this already contributes to T to the power 3 can I say it is as good as finding coefficient of T to the power 6 in this so T to the power 3 I dropped correct now this term the first term here is 1 minus 3 T to the power 6 other term would be 3 T to the power 12 now the moment you exceed the requirement you don't have to write those terms because they would anyways not contribute towards T to the power 6 okay so even though since you're doing it for the first time I'm writing all the terms down okay now guys this term this term is a infinitely big term there will be so many terms in this term remember if there is a negative integer power you had done this in the bridge course with me when I was discussing with you binomial expansion remember when n is a whole number it will have limited number of terms like these two guys but when n is not a whole number that means it's a negative integer or fraction you'll have infinitely many terms so do we sit and write those infinitely many terms the answer is no okay why the answer is no because see you want to generate T to the power 6 right how many ways can you generate T to the power 6 think about that if you pick a 1 from here if you pick a 1 from here and if you pick a term which is containing T to the power 6 here then T to the power 6 will be made so basically I'm thinking of all possibilities by which I can generate T to the power 6 correct second way is if I pull this term from here again if I pull this term from here and again I pick a constant term from here I will be able to generate T to the power 6 correct please note anyways these terms are not going to contribute towards T to the power 6 term that's why I was saying you don't have to even write it down so effectively speaking you have only these two expressions with you where this is infinitely many terms okay and this has got only two terms so what I want to say here is I'm looking for the constant term over here and that term which will give me T to the power 6 from here are you getting my point so these two if I am able to find out my job is done I don't have to go into and search in the entire ocean right so this is an ocean term so many terms are there infinitely many I don't have to write them all because I don't need them all what do I need I need the constant and why even I need the constant because constant into this will generate T to the power 6 because my ultimate problem statement is to generate T to the power 6 term and I need T to the power 6 because 1 into this term will generate T to the power 6 that's it there is no other requirement for you from this infinitely many term right now the question is sir here we are and constant is very easy that constant will be one in fact how do you get T to the power 6 coefficient that is the challenge so what I'm going to do is I'm going to link one more theory with this so see this problem is teaching you so many things so let's go to the side of the page and I'm going to now teach you how to find how to find coefficient of x to the power r in 1 minus x to the power n kind of a term as you can see this term is 1 minus T to the power minus n kind of a term so let me write in terms of the same variable else you'll get confused so how to find T to the power r in such kind of a term now see if you open this term in the bridge course we had seen it will be 1 plus minus n times minus T those who have not those who were not there with us in the bridge course please do ping me I will send you all the videos of the bridge course you can watch it at your free time by 2 factorial minus T square okay next term will be minus n minus n minus 1 minus n minus 2 minus T cube by 3 factorial and so on and so on note that it'll never end it'll go on and on for infinity yeah yeah yeah yeah we did this in the limits hurry and that's why i was saying you are all aware of it if you simplify this you will realize you are writing this miraculously all the negative signs are going to disappear okay dot dot dot so let's say if i go to let's say some T to the power rth term what would be the coefficient here that is what i'm looking for okay now if you look at the pattern you can see that if you're going to third term you are writing from n to n plus 2 and you're dividing by 3 factorial right so can i see if i'm going to T to the power r then the coefficient of T to the power r will be n n plus 1 n plus 2 all the way till n plus r minus 1 divided by r factorial correct so this is my answer but this doesn't look that great to me so what i'll do is i'll try to simplify this by multiplying and dividing with n minus 1 factorial because if i do my numerator will become n plus r minus 1 factorial by n minus 1 factorial r factorial correct and this term is nothing but n plus r minus 1 cr so i would like you to remember this formula because it's very very helpful okay please remember this because you're not going to derive all these things in the examination hollow so coefficient of T to the power r in 1 minus T to the power minus n is given by this formula please note this down okay now coming back to this question so if i want what is the coefficient of T to the power 6 so if i want to find what is the coefficient of T to the power 6 in 1 minus T to the power minus 4 okay how will i find it out remember my n is 4 my r is 6 so as per my formula this answer will be 4 plus 6 minus 1 c 6 which is nothing but 9 c 6 so this term here would be 9 c okay constant term anyways will be 1 i don't think so there's a problem with the constant term now let's conclude the question so in order to get T to the power 6 coefficient from this entire expression in order to get coefficient of T to the power 6 from this entire expression 1 should multiply with 9 c 6 and minus 3 should multiply with 1 there you go this is your answer so if you simplify this it becomes 9 factorial by 6 factorial 3 factorial minus 3 that's nothing but 9 into 8 into 7 by 6 minus 3 that's nothing but 81 okay so there are 81 ways so i'm going back to my question question was what are the number of solutions for this equation correct correct right under the constraint that x y z should be between 1 to 6 and there's no constraint on d other than it should be greater than equal to 0 so the number of ways to do this was 81 okay now my question was not this my question was in how many ways my question was i'm just coming back to my question my question was in how many ways can i get this to be greater than 9 when i'm rolling three dice isn't it so what did i say it is total cases minus those cases where x plus y plus z was less than equal to 9 this you've already figured out which was 81 total cases is 6 to the power 3 because the number of ways to roll your dice that is total number of cases to get outcome from them is 6 to the power 3 so your answer is 216 216 minus 81 how much will it give you i think it gives you 100 and uh uh how much 35 correct yes or no okay whoa this was a huge question right but this question gave me an opportunity to discuss two things in fact three things together one being how do you solve when you have to find something which is having greater than inequality and at the same time there's an upper restriction on the values of these variables and how do we use multinomial theorem to get this task done so this is a beautiful question beautiful question got this okay i understand this question the first time you're doing such a massive scale question is this fine can we begin with greater than equal to inequality see greater than inequality is like greater than 10 correct no aditya if you want to solve greater than equal to nine you just convert it to greater than equal greater than 10 solve it like that anyways you're dealing with integrals values no got it okay so uh i don't want you to take a complicated question but yes i will definitely like you to solve one small question on this very small question just to make sure that your multinomial theorem concept is understood well uh let me frame a question very simple question very simple question in how many ways can you distribute in how many ways can you distribute let's say i take three only distribute uh 15 blankets 15 blankets to three beggars how many ways can you distribute 15 blankets to three beggars such that no beggar gets more than four blanket oh sorry no beggar gets more than six blanket or let's say seven blanket question is simple in how many ways can you distribute 15 blankets to these beggars such that no beggar gets more than seven blanket try this out okay so even i'll help you to solve this see think as if this is what other values you can assign to x so basically i'm raising it to some power okay and there are three such beggars so you can basically do this correct in this if i'm able to find out the coefficient of t to the power 15 my job will be done guys let me clarify why i'm so keen about finding coefficient of t to the power 15 because coefficient keeps a count of how many times t to the power 15 is generated and the number of counts t to the power 15 is generated is the number of ways i will get a sum of 15 from x y z that is the reason why i'm doing it yeah i'm not solving it completely so i'm just giving you a hint so from here i need coefficient of t to the power 15 yeah after this i think you should be trying out on your own uh don't calculate your answer uh just give me an expression so that i know that uh you know you are on the right track oh yes yes yes yes sorry thanks for reminding me that okay guys let's do this so the first term will give you 1 minus 3 t to the power 8 now other term would be 3 t to the power 16 don't even write it why why you are writing t to the power 16 will it contribute to t to the power 15 it has already exceeded t to the power 15 okay but this guy is an ocean okay this guy is an ocean now for ocean from this ocean i need certain terms which terms let's decide see i want t to the power 15 so if i have one over here i would need t to the power 15 from here that is one of the terms i need so this is the term question mark i need i have t to the power 8 over here so i need i'm putting dot dot dot so i need some t to the power 7 term dot dot dot correct so these two terms one is your white question mark and other is your yellow question mark those two terms i need that's it i don't need the entire ocean don't waste time writing whole thing okay now if you want t to the power 15 this will be n plus r minus 1 c r that's nothing but 17 c 15 that is nothing but 17 c 2 so this guy will be 17 c 2 now see the beauty of the formula if you remember it in one shot you can get it you don't have to write down and then you know figure out the this thing similarly this guy this guy would be n plus r minus 1 c r that's 9 c 2 this guy is 9 c 2 okay so the answer would be listen to this answer would be 1 into 17 c 2 minus 3 into 9 c 2 this is going to be your answer is that clear is it coming out 28 aditya so simple let's check this is 17 into 16 by 2 17 into 8 this is 3 into 9 into 8 by 2 which is 4 so uh this is this is this is this is uh around 136 and this is around 12 into 9 which is 108 oh 28 ways absolutely fine absolutely correct is the idea clear question there sorry approach clear not a difficult thing if you master it no you will find that you are able to solve see multi-government is a bigger version of whatever you have learned okay it can be used to solve even the basic questions but that would be like you know you're picking a hammer to break an egg okay one question i would like to give you as a homework okay uh let's say there's a student homework question there's a student who has to attempt three papers paper one paper two paper three paper one has a maximum marks of 50 paper two has a maximum marks of 100 paper three has a maximum marks of again 100 okay in how many ways in how many ways can he score a total of a total of 175 marks from these three papers okay in how many ways can he score a total of 175 marks from these three papers okay so solve this as a homework question do let me know uh personally to me if you have solved it guys be careful he cannot get more than 100 150 in these papers right so don't try to distribute it like 75 here 50 here 50 here this is not going to be valid because he cannot score 75 in paper number one okay so there's an upper restriction on the values okay so solve this and let me know yes yes yes 0 is allowed 0 is allowed and he can only score integral marks there's no half a mark okay so he can get a zero mark 0 is allowed 0 is allowed but no fraction mark no fraction mark because most of you would be saying sir what if he gets 38 and a half there and you know 96.75 there no fraction marks only integral marks okay as the problem will become super complicated is this clear let's now take a break