How to find the sum of the given series: $\sum_{k=0}^{\min(n ,m)} \binom{n}{k} \binom{m}{k}$?

Question

I wonder whether it is possible to calculate the following sum that involves the Binomial coefficients

$\sum_{k=0}^{\min(n ,m)} \binom{n}{k} \binom{m}{k}$ .

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 2

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.