Verification of a Combinatorial Identity I have a challenge for you combinatorial mathematicians. Is anyone willing to verify the following combinatorial identity? 
$$\sum_{k=0}^n\sum_{j=0}^m\binom{m}{j}\binom{k-j}{p-j}\binom{p}{k-j}\binom{n+1}{k+m-j}$$ 
$$=\binom{n+1}{p}\binom{n+1}{m}-\binom{n+1-m}{p-m}\binom{p}{n+1-m}$$
Detailed steps or references would be appreciated. If this formula already resides in a long lost combinatorial book, I would love to know about it. 
 A: After browsing through several combinatorial books, it seems that the notion I was looking for is called a Vandermonde identity. Hence, I can now provide a very simple proof. 
We first note that $\binom{n+1-m}{p-m}\binom{p}{n+1-m}$ can be added to the summation, so that the indices range over all possible non-zero terms. Hence, disregarding the indices, we wish to show, 
$$\sum_{k}\sum_{j}\binom{m}{j}\binom{k-j}{p-j}\binom{p}{k-j}\binom{n+1}{k+m-j}=\binom{n+1}{p}\binom{n+1}{m}.$$
We perform a substitution of $l=k-j$ on the left side and then apply two Vandermonde identities [Riordan, pp. 9, 15],
$\displaystyle\sum_{k}\sum_{j}\binom{m}{j}\binom{k-j}{p-j}\binom{p}{k-j}\binom{n+1}{k+m-j}$
\begin{align}
\nonumber&=\sum_{l}\binom{p}{l}\binom{n+1}{m+l}\sum_{j}\binom{m}{j}\binom{l}{p-j} \\
\nonumber&=\sum_{l}\binom{p}{l}\binom{n+1}{m+l}\binom{m+l}{p} \\ 
\nonumber&=\sum_{l}\binom{p}{l-m}\binom{n+1}{l}\binom{l}{p} \\ 
\nonumber&=\sum_{l}\binom{l}{p}\binom{p}{l-m}\binom{n+1}{l} \\ 
\nonumber&=\binom{n+1}{p}\binom{n+1}{m}.  
\end{align}
