Find a closed-form expression for $\sum_{k=0}^n\left[\binom{n}{k}\binom{m}{k}\right]$, where $m$ is fixed. Find a closed-form expression for $\sum_{k=0}^n\left[\binom{n}{k}\binom{m}{k}\right]$, where $m$ is fixed.

That sigma is messing me up!  How would I simplify this equation algebraically(I prefer solving it using algebra)?  If there isn't a way to solve it algebraically, could you provide a counting solution?  Thanks in advance!
 A: Notice $\displaystyle\;\binom{m}{k} = 0$ whenever $k > m$. WOLOG, we only need to study the case $m \ge n$.
Given any polynomial $f(x) = \sum\limits_{k=0}^{\deg f} a_k x^k$, we will use $[x^k] f(x)$ to denote $a_k$, the coefficient of $x^k$ in $f(x)$.
By Binomial theorem, we have
$$\binom{n}{k} = [x^k] (1+x)^n\quad\text{ and }\quad
\binom{m}{k} = \binom{m}{m-k} = [x^{m-k}] (1+x)^m$$
This implies
$$\sum_{k=0}^n \binom{n}{k}\binom{m}{k}
= \sum_{k=0}^n \left([x^k](1+x)^n\right)\left([x^{m-k}](1+x)^m\right)
= [x^m] (1+x)^{n+m} = \binom{n+m}{m}
$$
A: Though you already have a nice algebraic answer, I'd like to offer a combinatorial answer, because I like them, and I hope you could appreciate them more by having one for this answer.
(For convention, I'll let $[n]=\{1,2...n\}$.) The left side counts the number of pairs $(A,B)$ where $A\subseteq [n], B\subseteq [m], |A|=|B|$. The right side counts size $m$ subsets of $[n+m]$. Now we partition $[n+m]$ into $[n]$ and $\{n+1, n+2...n+m\}$ (let's call it $[n+1,n+m]$), with the two sets having sizes $n$ and $m$ respectively.  Now, for each pair $(A,B)$ we can interpret $B$ as a subset of $[n+1,n+m]$ by identifying each $i\in B$ with $n+i\in [n+1,n+m]$ (this is clearly a bijection).
What we want now is a bijection between pairs $(A,B)$ with $A\subseteq [n]$, $B\subseteq[n+1,n+m]$, $|A|=|B|,$ and size $m$ subsets of $[n+m]$. Consider the set $[m]$: for a pair $(A,B)$, we look at $([m]- B)\cup A$. That is, we start with just the numbers greater than $n$, but then replace some of them with some numbers less than or equal to $n$, hence why we need $|A|=|B|$. Showing this to be a bijection is fairly straightforward, as each size $m$ subset of $[n+m]$ can be determined, and uniquely so, by which elements less than or equal to $n$ that is has, and which elements greater than $n$ that it's missing.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Approach $\ds{\mathbf{0}}$:
\begin{align}
\pars{1 + x}^{m}\pars{1 + x}^{n} & = \pars{1 + x}^{m + n}
\\[5mm] 
\sum_{i = 0}^{m}\sum_{j = 0}^{n}{m \choose i}{n \choose j}x^{i + j} & = 
\sum_{k = 0}^{m + n}{m + n \choose k}x^{k}
\\[5mm] 
\sum_{k = 0}^{m + n}\bracks{\sum_{i = 0}^{m}{m \choose i}{n \choose k - i}
\bracks{k - n \leq i \leq k}}x^{k} & = 
\sum_{k = 0}^{m + n}{m + n \choose k}x^{k}
\end{align}

Just compare coefficients of $\ds{x^{n}}$ in both sides
  $\pars{~\mbox{as}\ \bbox[#ddd,4px]{\texttt{@achile hui}}\ \mbox{did it}~}$. 


Approach $\ds{\mathbf{1}}$:
\begin{align}
\color{#f00}{\sum_{k = 0}^{n}{n \choose k}{m \choose k}} & =
\sum_{k = 0}^{\infty}{n \choose k}{m \choose m - k} =
\sum_{k = 0}^{\infty}{n \choose k}\
\overbrace{\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{m} \over z^{m - k + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{m \choose m - k}}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{m} \over z^{m + 1}}
\sum_{k = 0}^{\infty}{n \choose k}z^{k}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{m} \over z^{m + 1}}
\pars{1 + z}^{n}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{m + n} \over z^{m + 1}}
\,{\dd z \over 2\pi\ic} = \color{#f00}{m + n \choose m}
\end{align}
