Re: another partition question

On Oct 8, 3:00 am, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article
<d37b9f70-97ea-4ed2-8423-caff50d06...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

nikl...@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 (ge...@xxxxxxxxxxxxxxx) (i -> u for email)

Thanks gerry for the answer

Any suggestions about
b)

If it had be positive partitions i could have used p(n,k) =p(n-1,k-1 )
+ p(n-k,k) but when there is 0 allowed , i get something like p(n,k) =
p(n,k-1)(add a 0 ) + p (n-1,k) (u can increase one ofthe terms by 1 ).
Now when increasing any of the terms by 1, we are over computing
because the element +1 needn't be in set. Any suggestions how to
remove overcounting or a better recurrence?


c)
Can the same problem be solved combinatorially in a simpler manner ?
.

Quantcast