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

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

satellite73Best ResponseYou've already chosen the best response.0
can't wait to see this
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.2
http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
 2 years ago

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

satellite73Best ResponseYou've already chosen the best response.0
how cool is that? stars and bars, nice visual!
 2 years ago

shaan_iitkBest ResponseYou've already chosen the best response.0
One 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!)
 2 years ago

FoolForMathBest 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.
 2 years ago

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

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

shaan_iitkBest ResponseYou've already chosen the best response.0
okk .. 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..
 2 years ago
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