TITLE: How to prove combinatorial identity $\sum_{k=0}^s{s\choose k}{m\choose k}{k\choose m-s}={2s\choose s}{s\choose m-s}$?
QUESTION [7 upvotes]: The following combinatorial identity have been verified via maple, but I can not prove it.
Who can prove it without WZ mehtod? 
$$\sum_{k=0}^s{s\choose k}{m\choose k}{k\choose m-s}={2s\choose s}{s\choose m-s}$$.

REPLY [2 votes]: By  way of  enrichment here  is  another algebraic  proof using  basic
complex variables.

Suppose we are trying to show that
$$\sum_{k=0}^n {n\choose k} {m\choose k} {k\choose m-n}
= {2n\choose n}{n\choose m-n}.$$
Introduce the integral representations 
$${m\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{k+1}} \; dz
\quad\text{and}\quad
{k\choose m-n} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^k}{w^{m-n+1}} \; dw.$$
where $m\ge n.$
This gives for the sum the representation
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{m-n+1}}
\sum_{k=0}^n {n\choose k} \left(\frac{1+w}{z}\right)^k
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{m-n+1}}
\frac{(1+w+z)^n}{z^n}
\; dw \; dz.$$
Extracting the inner residue we obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} 
{n\choose m-n} (1+z)^{2n-m} \; dz
\\ = {n\choose m-n} 
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n}}{z^{n+1}} \; dz
\\ = {2n\choose n} {n\choose m-n}.$$
We have  not made use of  the properties of complex  integrals here so
this  computation  can  also   be  presented  using  just  algebra  of
generating functions.

There is a similar but somewhat more advanced calculation at this
MSE link. 

Apparently  this method is  due to  Egorychev although  some of  it is
probably folklore.