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AravindG
 3 years ago
find the no: of integral solutions of inequation x+y+z+u<=65
AravindG
 3 years ago
find the no: of integral solutions of inequation x+y+z+u<=65

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FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.2I can do this using binomial coefficients.

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.2Using stars and bars, the answer is given by \[ \huge \sum \limits_{n=0}^{65} \binom{n+3}{3} \]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0can't wait to see this

moneybird
 3 years ago
Best ResponseYou've already chosen the best response.2this question is like putting three bookmarks in a book

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0how cool is that? stars and bars, nice visual!

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0One can also think logically and get the solution .. say you have x+y+z+u <= k (some integer) and you have to find the number of solutions then it is like finding the number of ways where k sweets can be distributed to 4 kids and total sweets distributed may not be equal to k. Now lets imagine a situation where we have 4 women and k men. How many ways of arranging them? (4+k)!/(4!*k!) Now say I say that all men who come to the left of any woman are given to that woman (not literally) .. then we have solved our distribution problem and hence our equation.. so I believe the answer should be 69!/(4!*65!)

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.2@shaan_iitk How do you think stars and bars work? That is exactly the same thing you elaborated.

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry but I am not aware of stars and bars ..

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.2I have posted the url before.

shaan_iitk
 3 years ago
Best ResponseYou've already chosen the best response.0okk .. I didn't look at that.. I only looked at your expression which was in summation form.. I thought logically we don't need that summation..
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