How to prove combinatorial identity $\sum_{k=0}^s{s\choose k}{m\choose k}{k\choose m-s}={2s\choose s}{s\choose m-s}$? The following combinatorial identity have been verified via maple, but I can not prove it.
Who can prove it without WZ mehtod? 
$$\sum_{k=0}^s{s\choose k}{m\choose k}{k\choose m-s}={2s\choose s}{s\choose m-s}$$.
 A: Use subset-of-a-subset**, isolate non-index term, and then use Vandermonde*** to arrive at the result.
$$\begin{align}
\sum_{k=0}^s \color{blue}{\binom sk}\binom mk \color{blue}{\binom k{m-s}} &=\sum_{k=0}^{s}\color{blue}{\binom sk{\binom k{m-s}}} \binom mk \\
&=\sum_{k=0}^{s}\color{blue}{\binom s{m-s}\binom{2s-m}{k-m+s}}\binom mk\\
&=\binom s{m-s}\sum_{k=0}^{s}\binom{2s-m}{\color{red}k-m+s}\binom m{m\color{red}{-k}}\\
&=\binom s{m-s}\binom{2s}s \\&=\binom{2s}s\binom s{m-s}\qquad \blacksquare
\end{align}$$

** Subset-of-a-subset:
$$\color{blue}{\binom ab\binom bc=\binom ac \binom {a-c}{b-c}}$$
*** Vandermonde:
$$\sum_{r=0}^{a-b} \binom a{\color{red}r+b} \binom c{d\color{red}{-r}}=\binom {a+c}{b+d}$$
A: Let $n=m-s$; we may assume that $0\le n\le s$, as otherwise both sides are zero. The righthand side is the number of ordered pairs $\langle A,B\rangle$ such that $A$ is an $s$-subset of $[2s]$ and $B$ is an $m$-subset of $[s]$. 
So is the lefthand side, though this is less obvious. Note that the $k$ term is zero unless $n\le k\le s$. For those $k$ we first pick $k$ elements of $[s]$ and then pick $n$ of those elements to be $B$. This leaves $k-n$ of the chosen elements unused; they will be $A\cap([s]\setminus B)$. The remaining 
$$s-(k-n)=s+n-k=m-k$$ 
elements of $A$ must come from $B\cup([2s]\setminus[s])$, a set that has $s+n=m$ elements. Altogether there are
$$\binom{s}k\binom{k}n\binom{m}{m-k}=\binom{s}k\binom{k}n\binom{m}{k}$$
to complete the choice of $A$ and $B$ for this value of $k$, and summing over $k$ now yields the result.
A: 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.
