TITLE: Evaluate $\sum_{k=0}^{n} {n \choose k}{m \choose k}$ for a given $n$ and $m$.
QUESTION [12 upvotes]: How do I evaluate $\sum_{k=0}^{n} {n \choose k}{m \choose k}$ for a given $n$ and $m$.
I have tried to use binomial expansion and combine factorials, but I have gotten nowhere. I don't really know how to start this problem.
The answer is ${n+m \choose n}$. Any help is greatly appreciated.
EDIT: I'm looking for  a proof of this identity.

REPLY [7 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{k=0}^{n}{n \choose k}{m \choose k}:\ {\large ?}}$

${\large\tt\mbox{Hereafter, I'll illustrate a general method:}}$

\begin{align}&\color{#66f}{\large\sum_{k=0}^{n}{n \choose k}{m \choose k}}
=\sum_{k=0}^{n}{n \choose k}\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{m} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{m} \over z}\,
\sum_{k=0}^{n}{n \choose k}\pars{1 \over z}^{k}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{m} \over z}\,
\pars{1 + {1 \over z}}^{n}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{m + n} \over z^{n + 1}}\,
\,{\dd z \over 2\pi\ic} = \color{#66f}{\large{m + n \choose n}}
\end{align}