TITLE: Proof of the identity $\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}$
QUESTION [15 upvotes]: Prove the identity:
$$\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}.$$

REPLY [8 votes]: $$
\begin{align}
\sum_k\binom{p}{k}\binom{q}{k}\binom{n+k}{p+q}
&=\sum_{j,k}\binom{p}{k}\binom{q}{k}\binom{n}{p+q-j}\binom{k}{j}\tag{1}\\
&=\sum_{j,k}\binom{p}{k}\binom{n}{p+q-j}\binom{q}{j}\binom{q-j}{q-k}\tag{2}\\
&=\sum_{j}\binom{p+q-j}{q}\binom{n}{p+q-j}\binom{q}{j}\tag{3}\\
&=\sum_{j}\binom{n-q}{n-p-q+j}\binom{n}{q}\binom{q}{j}\tag{4}\\
&=\sum_{j}\binom{n-q}{p-j}\binom{n}{q}\binom{q}{j}\tag{5}\\
&=\binom{n}{p}\binom{n}{q}\tag{6}
\end{align}
$$
Explanation
$(1)$ $\displaystyle\binom{n+k}{p+q}=\sum_j\binom{n}{p+q-j}\binom{k}{j}$
$(2)$ $\displaystyle\binom{q}{k}\binom{k}{j}=\binom{q}{j}\binom{q-j}{q-k}$
$(3)$ $\displaystyle\sum_k\binom{p}{k}\binom{q-j}{q-k}=\binom{p+q-j}{q}$
$(4)$ $\displaystyle\binom{p+q-j}{q}\binom{n}{p+q-j}=\binom{n-q}{n-p-q+j}\binom{n}{q}$
$(5)$ $\displaystyle\binom{n-q}{n-p-q+j}=\binom{n-q}{p-j}$
$(6)$ $\displaystyle\sum_j\binom{n-q}{p-j}\binom{q}{j}=\binom{n}{p}$