Re: another partition question

In article
<d37b9f70-97ea-4ed2-8423-caff50d06289@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
niklaus@xxxxxxxxx wrote:

how many sols we have for a + b +c + d + e = n 0<=a<=k, 0<=b<=k
same bounds for all the elements ?

This will be the coefficient of x^n in p(x) = (1 + x + ... + x^k)^5.
Can you see why?
Now p(x) = (1 - x^{k+1})^5 (1 - x)^{-5}
and you can use the binomial theorem on each of the two factors
and then pick out the coefficient of x^n in the product of the
two expansions.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.