Identity involving product of two binomial coefficients Emprically it looks like the following identity holds, but I haven't been able to prove it. Can anyone find a proof?
$$
\binom{m+k}{k}\binom{n+k}{k}=\sum_{i\geq0}\binom{m}{i}\binom{n}{i}\binom{m+n+k-i}{k-i}
$$
For some context, a corollary would be that the generating function for the LHS keeping $m$ and $n$ fixed is
$$
\sum_{k\geq0}\binom{m+k}{k}\binom{n+k}{k}x^k=\frac{\sum_{i\geq0}\binom{m}{i}\binom{n}{i}x^i}{(1-x)^{m+n+1}}
$$
 A: We seek to prove that
$${m+k\choose k} {n+k\choose k}
= \sum_{q\ge 0} {m\choose q} {n\choose q} {m+n+k-q\choose k-q}.$$
We start on the RHS with
$$\sum_{q\ge 0} {m\choose q}
{n\choose n-q} {m+n+k-q\choose k-q}
\\ = [z^n] (1+z)^n [w^k] (1+w)^{m+n+k}
 \sum_{q\ge 0} {m\choose q} z^q w^q (1+w)^{-q}
\\ = [z^n] (1+z)^n [w^k] (1+w)^{m+n+k}
\left(1+\frac{zw}{1+w}\right)^m
\\ = [z^n] (1+z)^n [w^k] (1+w)^{n+k}
(1+w+zw)^m
\\ = [w^k] (1+w)^{n+k} [z^n] (1+z)^n
(1+w(1+z))^m
\\ = [w^k] (1+w)^{n+k} [z^n] (1+z)^n
\sum_{q=0}^m {m\choose q} w^q (1+z)^q
\\ = [w^k] (1+w)^{n+k}
\sum_{q=0}^m {m\choose q} {n+q\choose q} w^q
\\ = \sum_{q=0}^k {m\choose q} {n+q\choose q}
{n+k\choose k-q}.$$
Observe that
$${n+q\choose q} {n+k\choose k-q}
= \frac{(n+k)!}{q!\times n!\times (k-q)!}
= {n+k\choose k} {k\choose q}.$$
We get
$${n+k\choose k} \sum_{q=0}^k {m\choose q}
{k\choose q}
= {n+k\choose k} \sum_{q=0}^k {m\choose q}
{k\choose k-q}
\\ = {n+k\choose k} [z^k] (1+z)^k
\sum_{q=0}^k {m\choose q} z^q.$$
We may  extend $q$ to  infinity owing  to the coeffcient  extractor in
$z$:
$${n+k\choose k} [z^k] (1+z)^k
\sum_{q\ge 0} {m\choose q} z^q
\\ = {n+k\choose k} [z^k] (1+z)^{m+k}
= {n+k\choose k} {m+k\choose k}.$$
This is the claim.
A: That is known as Suranyi's formula, and you can find
a demonstration in this paper.
Also interesting is the context in which it is analyzed
in this other paper.
