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How can we find the sum of the following series

$$\sum_{i=0}^p \binom{m-q+1+i}{i} \binom{n+q-1-i}{n-i}=\sum_{i=0}^p\frac{(m-q+1-i)!}{ i! (m-q+1)!}\frac{ ( n + q-1-i)!}{ (q-1)! (n-i)!}$$ where $p < n,m$?

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Do you denote by $!n$ the factorial? Usually it is denoted by $n!$. – martini Dec 6 '12 at 14:21
$!n$ is not the notation for factorial, but rather for derangements. – Cameron Buie Dec 6 '12 at 14:30

Here is an answer computed by maple

$${n+q-1\choose n}{_2F_1(-n,m-q+2;\,-n-q+1;\,1)}-{m-q+2+p\choose p+1}{n+q -2-p\choose n-p-1}$$ $$\times\,{_3F_2(1,-n+1+p,m-q+3+p;\,p+2,-n-q+2+p;\,1)} $$

where $_2F_1$ and $_3F_2$ are the hypergeometric function.

The same answer can be computed by Mathematica $9$.

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If $p\ge n$, then $$ \begin{align} \sum_{i=0}^p\binom{m-q+1+i}{i}\binom{n+q-1-i}{n-i} &=\sum_{i=0}^p\binom{m-q+1+i}{m-q+1}\binom{n+q-1-i}{q-1}\\ &=\binom{m+n+1}{m+1} \end{align} $$ However, if $p\lt n$ I don't think there is a closed form (that, in general, doesn't involve hypergeometric functions).

To confirm Mhenni Benghorbal, Mathematica 8 gives $$ \binom{n+q-1}{n}\,_2F_1(-n,m-q+2;-n-q+1;1) -\binom{m+p-q+2}{p+1}\binom{n-p+q-2}{n-p-1}\\ \times\,_3F_2(1,-n+p+1,m+p-q+3;p+2,-n+p-q+2;1) $$

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