How to find the sum of the given series: $\sum_{k=0}^{\min(n ,m)} \binom{n}{k} \binom{m}{k}$?  I wonder whether it is possible to calculate the following sum that involves the Binomial coefficients 
$\sum_{k=0}^{\min(n ,m)} \binom{n}{k} \binom{m}{k}$ .
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k=0}^{\min\braces{n,m}}{n \choose k}{m \choose k} & =
\sum_{k=0}^{\infty}{n \choose k}{m \choose m - k} \\[5mm] & =
\sum_{k=0}^{\infty}{n \choose k}
\bracks{z^{m - k}}\pars{1 + z}^{m}
\\[5mm] & =
\bracks{z^{m}}\pars{1 + z}^{m}
\sum_{k=0}^{\infty}{n \choose k}z^{k}
\\[5mm] & =
\bracks{z^{m}}\pars{1 + z}^{m + n} =
\bbx{m + n \choose m}
\end{align}
A: Yes, it is in fact possible to sum this. The answer is
$$\sum_{k=0}^n\binom{n}{k}\binom{m}{k}=\binom{m+n}{n}$$
assuming that $n\leq m$. This comes from the fact that
$$\sum_{k=0}^n\binom{n}{k}\binom{m}{k}=\frac{m+n}{n}\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{m}{k}$$
The way that I solved this problem was using WZ theory to find a recursion for the sums, which is quite algebraically intensive, and I don't know if there is a better way of showing this recursion.
