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The problem asks us to calculate:
$$ \sum_{i = 0}^{n}(-1)^i \binom{n}{i} \binom{n}{n-i}$$

The way I tried solving is:

The given sum is the coefficient of $x^n$ in $ (1+x)^n(1-x)^n $, which is $ (1 - x^2)^n $. The coefficient of $x^n$ in $(1 -x^2)^n$ is $$ (-1)^{n/2} \binom{n}{n/2}. $$

Am I doing it right?

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Thanks, fixed it. – user813 Sep 22 '10 at 15:57
up vote 4 down vote accepted

Your solution is correct for even $n$.

If $n$ is odd then your last sentence should read "The coefficient of $x^n$ in $(1-x^2)^n$ is $0$." This is because only even powers of $x$ occur when expanding $(1-x^2)^n$.

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Ah, good. Thanks. – user813 Sep 22 '10 at 16:01

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} &\color{#f00}{\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{n \choose n - k}} = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\ \overbrace{% \oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{n} \over z^{n - k + 1}} \,{\dd z \over 2\pi\ic}}^{\ds{=\ {n \choose n - k}}} \\[3mm] = &\ \oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{n} \over z^{n + 1}}\ \overbrace{\sum_{k = 0}^{n}{n \choose k}\pars{-z}^{k}} ^{\ds{=\ \pars{1 - z}^{n}}}\ \,{\dd z \over 2\pi\ic} = \oint_{\verts{z} = 1^{-}}{\pars{1 - z^{2}}^{n} \over z^{n + 1}} \,{\dd z \over 2\pi\ic} \\[3mm] = &\ \sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k}\ \underbrace{% \oint_{\verts{z} = 1^{-}}{1 \over z^{n - 2k + 1}}}_{\ds{=\ \delta_{n,2k}}} \,{\dd z \over 2\pi\ic} = \color{#f00}{\left\lbrace\begin{array}{lcl} \ds{\pars{-1}^{n/2}{n \choose n/2}} & \mbox{if} & \ds{n}\ \mbox{is}\ even \\[2mm] \ds{0} && \mbox{otherwise} \end{array}\right.} \end{align}

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