Show this combinatorical identity

Question

$$\sum_{k=0}^m {n \choose k} {n-k \choose m-k} = 2^m {n \choose m}, m

I know that $2^m$ represents the number of subsets of a set of length $m$, which I can see there being a connection to the ${n \choose k}$ term, but I can't see how the combination it's multiplied by affects this.

Answer 1

A "double-counting" proof

Let $m

First, we can select the $m$ objects to be painted ($\binom{n}{m}$ ways), and then the subset to be placed inside the box ($2^m$ waks). Multiplicative principle gives us

$$ \binom{n}{m}2^m $$

to do this. Another way is first select the number of objets we want to put on the box, $k$ (of course, $0\leq k \leq m$), then select the box-objects ($\binom{n}{k}$ ways), and then the rest of painted objects ($m-k$ object), in $\binom{n-k}{m-k}$ ways. This way, we have

$$ \binom{n}{k} \binom{n-k}{m-k}$$

ways to do this, and adding the (disjoint) cases for $k=0, 1, \dots, m$ we have

$$ \sum_{k=0}^n \binom{n}{k} \binom{n-k}{m-k}$$

options. We counted the same number in two different ways, so they have to be equal:

$$ \sum_{k=0}^n \binom{n}{k} \binom{n-k}{m-k} = 2^m \binom{n}{m} $$

An algebraic proof

$$\begin{align*} & \sum_{k=0}^m \binom{n}{k} \binom{n-k}{m-k} \\ =& \sum_{k=0}^m \frac{n!}{k! (n-k)!}\cdot \frac{(n-k)!}{(m-k)!(n-m)!} \\ =& \sum_{k=0}^m \frac{n!}{k!(m-k)!(n-m)!}\\ =& \sum_{k=0}^m \frac{n!}{(n-m)!m!}\cdot \frac{m!}{k!(m-k)!} \\ =& \sum_{k=0}^m \binom{n}{m} \binom{m}{k} \\ =& \binom{n}{m} \sum_{k=0}^m \binom{m}{k} \\ =& \binom{n}{m} \cdot 2^m \end{align*} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

An analytical approach:

\begin{align} \sum_{k = 0}^{m}{n \choose k}{n - k \choose m - k} & = \bracks{z^{m}}\sum_{\ell = 0}^{\infty}z^{\ell} \bracks{\sum_{k = 0}^{\ell}{n \choose k}{n - k \choose \ell - k}} = \bracks{z^{m}}\sum_{k = 0}^{\infty}{n \choose k} \sum_{\ell = k}^{\infty}{n - k \choose \ell - k}z^{\ell} \\[5mm] & = \bracks{z^{m}}\sum_{k = 0}^{\infty}{n \choose k} \sum_{\ell = 0}^{\infty}{n - k \choose \ell}z^{\ell + k} = \bracks{z^{m}}\sum_{k = 0}^{\infty}{n \choose k}z^{k} \sum_{\ell = 0}^{\infty}{n - k \choose \ell}z^{\ell} \\[5mm] & = \bracks{z^{m}}\sum_{k = 0}^{\infty}{n \choose k}z^{k}\pars{1 + z}^{n - k} = \bracks{z^{m}}\bracks{\pars{1 + z}^{n} \sum_{k = 0}^{\infty}{n \choose k}\pars{z \over 1 + z}^{k}} \\[5mm] & = \bracks{z^{m}}\bracks{\pars{1 + z}^{n}\pars{1 + {z \over 1 + z}}^{n}} = \bracks{z^{m}}\pars{1 + 2z}^{n} = \bracks{z^{m}}\sum_{\ell = 0}^{n}{n \choose \ell}\pars{2z}^{\ell} \\[5mm] & = \bbx{\ds{{n \choose m}2^{m}}} \end{align}