prove $\sum_{k = 0}^{n} \binom{n}{k} \binom{m-n}{n-k} = \binom{m}{n}$ prove $\sum_{k = 0}^{n} \binom{n}{k} \binom{m-n}{n-k} = \binom{m}{n}$
Attempt:I was thinking of trying to prove this through induction, but I am having trouble with a base case:
base case:  let $n = 2$: $$LHS = \binom{2}{0} \binom{m-2}{2} + \binom{2}{1} \binom{m-2}{1} + \binom{2}{2} \binom{m-2}{0} \\ = \frac{(m-2)[(m-3) + 2!2!(m-3)!] + 2!}{2!} \ (after\ simplification)$$
$$RHS = \frac{m!}{2!(m-2)!}$$
But I am stuck as what to try next to at least equate these two expressions.
Note: I took a look at Vandermonde's identity on wikipedia and the ensuing proof, but the proof still leaves out how to make the transition between the two initial expressions
 A: Use Pascal’s identity a few times:
$$\begin{align*}
\binom20\binom{m-2}2+\binom21\binom{m-2}1+\binom22\binom{m-2}0&=\binom{m-2}2+2\binom{m-2}1+\binom{m-2}0\\
&=\left(\binom{m-2}2+\binom{m-2}1\right)+\\
&\qquad\qquad\left(\binom{m-2}1+\binom{m-2}0\right)\\
&=\binom{m-1}2+\binom{m-1}1\\
&=\binom{m}2\;.
\end{align*}$$
But why start at $n=2$? Starting at $n=0$ makes the base case very simple.
A: A much easier approach would be to look at both sides as the coefficient of $x^n$ in the binomial expansion of $(1+x)^m$. 
The right hand side follows immediately. 
For the left hand side, write $(1+x)^m$ as $(1+x)^n$$(1+x)^{m-n}$ and use the binomial theorem for each term in the product individually. 
The coefficient for $x^n$ in this case is precisely the expression on the left hand side. 
A: Combinatorics proof: From a group of $m$ people (among which $n$ are left-handed) we can pick a committee of $n$ people by picking $k$ left-handers and $n-k$ right-handers, $0\le k\le n$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\color{#f00}{\sum_{k = 0}^{n}{n \choose k}{m - n \choose n - k}} & =
\sum_{k = 0}^{n}{n \choose k}\
\overbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{m - n} \over z^{n - k + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{{m - n \choose n - k}}}\ =\
\oint_{\verts{z} = 1}{\pars{1 + z}^{m - n} \over z^{n + 1}}\
\overbrace{\sum_{k = 0}^{n}{n \choose k}z^{k}}^{\ds{\pars{1 + z}^{n}}}\
\,{\dd z \over 2\pi\ic}
\\[3mm] & =\ \underbrace{%
\oint_{\verts{z} = 1}{\pars{1 + z}^{m} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic}}_{\ds{{m \choose n}}}\ =\
\color{#f00}{{m \choose n}}
\end{align}
