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How can I simplify the following expression?

$$\sum_{j=0}^{k} \binom{n-j}{p} \binom{m+j}{q}$$

where $n,m,p,q,k$ are positive constants such that $n-k \ge p$ and $m \ge q$.

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2  
You'll need to tell us what $k$ is, too :) –  Thomas Andrews Dec 3 '12 at 18:24
1  
As it stands, you cannot simplify it to a single expression. –  Phira Dec 3 '12 at 18:28
    
I assume that $k=n-p$. In any case, try generating series. –  Eric Naslund Dec 3 '12 at 18:30
    
Yeah, I updated the condition for k. –  Shashwat Kumar Dec 3 '12 at 18:32
    
I tried to make few polynomial expansions to achieve the above form but all in vain. –  Shashwat Kumar Dec 3 '12 at 18:36

2 Answers 2

If $n-k=p$ and $m=q$, then we can use the identity $$ \begin{align} (1-x)^{-p-1} &=\sum_{i=0}^\infty(-1)^i\binom{-p-1}{i}x^i\\ &=\sum_{i=0}^\infty\binom{p+i}{i}x^i\\ &=\sum_{i=0}^\infty\binom{p+i}{p}x^i\tag{1} \end{align} $$ to get $$ \begin{align} (1-x)^{-p-1}(1-x)^{-q-1} &=\sum_{i=0}^\infty\binom{p+i}{p}x^i\;\;\sum_{j=0}^\infty\binom{q+j}{q}x^j\\ &=\sum_{j=0}^\infty\sum_{i=0}^\infty\binom{p+i}{p}x^i\binom{q+j}{q}x^j\\ &=\sum_{j=0}^\infty\sum_{n=j+p}^\infty\binom{n-j}{p}x^{n-j-p}\binom{q+j}{q}x^j\\ &=\sum_{n=p}^\infty\sum_{j=0}^{n-p}\binom{n-j}{p}\binom{q+j}{q}x^{n-p}\tag{2} \end{align} $$ Of course, from $(1)$ we get $$ (1-x)^{-p-q-2}=\sum_{k=0}^\infty\binom{p+q+1+k}{p+q+1}x^k\tag{3} $$ Equating the terms with identical powers of $x$ and remembering that $n-k=p$ and $m=q$, we get $$ \sum_{j=0}^{n-p}\binom{n-j}{p}\binom{m+j}{q}=\binom{n+m+1}{p+q+1}\tag{4} $$

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Thanks for the solution. I think the general term for OP is unsolvable. –  Shashwat Kumar Dec 4 '12 at 8:44

Another hypergeometric expression from Maple:

$$ {n\choose p}{m\choose q}{\mbox{$_3$F$_2$}(1,m+1,-n+p;\,-n,m-q+1;\,1)}- {n-1-k\choose p}{m+1+k\choose q} {\mbox{$_3$F$_2$}(1,m+2+k,-n+1+k+p;\,-n+k+1,m+k+2-q;\,1)} $$

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