TITLE: How to find$\sum_{i,j,k\in \mathbb{Z}}\binom{n}{i+j}\binom{n}{j+k}\binom{n}{i+k}$ for $n \in \mathbb{N}$
QUESTION [6 upvotes]: Yeah, it's
$$\sum_{i,j,k\in \mathbb{Z}}\binom{n}{i+j}\binom{n}{j+k}\binom{n}{i+k}$$
and we are summing over all possible triplets of integers. It appears quite obvious that result is not an infinity. I tried to calculate
$$\sum_{i,j \in \mathbb{Z}} \binom{n}{i+j}\sum_{k \in \mathbb{Z}} \binom{n}{i+k} \binom{n}{j+k}$$
but it doesn't get easier for me. My intuition says that we are calculating something over set of size $3n$, but i couldn't get any idea right.
I'd appreciate some help on this super old exam task.

REPLY [5 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{equation}
\mathbf{\mbox{The Question:}\quad}
\sum_{j,k,\ell\ \in\ \mathbb{Z}}\,\,
{n \choose j + k}{n \choose k + \ell}{n \choose j + \ell} =\ ?\tag{1}
\end{equation}


\begin{equation}\mbox{Note that}\
\sum_{j,k,\ell\ \in\ \mathbb{Z}}\,\,
{n \choose j + k}{n \choose k + \ell}{n \choose j + \ell}=
\sum_{j,k\ \in\ \mathbb{Z}}\,\,
{n \choose j + k}\
\overbrace{\sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}
{n \choose j + \ell}}^{\ds{\equiv\ \,\mathcal{I}}}\tag{2}
\end{equation}

\begin{align}
\fbox{$\ds{\ \,\mathcal{I}\ }$} & =
\sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}{n \choose j + \ell} =
\sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}{n \choose n - j - \ell}
\\[4mm] & =
\sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{n} \over z^{n - j - \ell + 1}}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{n} \over z^{n - j + 1}}
\sum_{\ell\ \in\ \mathbb{Z}}{n \choose k + \ell}z^{\ell}\,{\dd z \over 2\pi\ic}
\\[4mm] & =
\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{n} \over z^{n - j + k + 1}}
\sum_{\ell\ \in\ \mathbb{Z}}{n \choose \ell}z^{\ell}\,{\dd z \over 2\pi\ic}
\\[4mm] & =
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{n} \over z^{n - j + k + 1}}
\pars{1 + z}^{n}\,{\dd z \over 2\pi\ic} = {2n \choose n - j + k} =
{2n \choose n + j - k}
\\[4mm] & =
\fbox{$\ds{\ \oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{2n} \over z^{n + j - k + 1}}\,{\dd z \over 2\pi\ic}\ }$} =
\fbox{$\ds{\ \,\mathcal{I}\ }$}
\end{align}

The original summation $\ds{\pars{1}}$ is reduced to $\pars{~\mbox{see expression}\ \pars{2}~}$:
\begin{align}
&\color{#f00}{\sum_{j,k,\ell\ \in\ \mathbb{Z}}\,\,
{n \choose j + k}{n \choose k + \ell}{n \choose j + \ell}} =
\sum_{j,k\ \in\ \mathbb{Z}}{n \choose j + k}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{2n} \over z^{n + j - k + 1}}\,{\dd z \over 2\pi\ic}
\\[4mm] = &\
\sum_{j\ \in\ \mathbb{Z}}\,\,\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{2n} \over z^{n + j + 1}}
\sum_{k\ \in\ \mathbb{Z}}{n \choose j + k}z^{k}\,{\dd z \over 2\pi\ic} =
\sum_{j\ \in\ \mathbb{Z}}\,\,\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{2n} \over z^{n + 2j + 1}}
\sum_{k\ \in\ \mathbb{Z}}{n \choose k}z^{k}\,{\dd z \over 2\pi\ic}
\\[4mm] = &\
\sum_{j\ \in\ \mathbb{Z}}\,\,\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{3n} \over z^{n + 2j + 1}}\,{\dd z \over 2\pi\ic} =
\sum_{j\ \in\ \mathbb{Z}}{3n \choose n + 2j} =
\sum_{j\ \in\ \mathbb{Z}}{3n \choose n + j}\,{1 + \pars{-1}^{j} \over 2}
\\[4mm] = &\
\half\sum_{j\ \in\ \mathbb{Z}}{3n \choose j} +
\half\,\pars{-1}^{n}\sum_{j\ \in\ \mathbb{Z}}{3n \choose j}\pars{-1}^{j} =
\half\,\pars{1 + 1}^{3n} + \half\,\pars{-1}^{n}\pars{1 - 1}^{3n}
\\[4mm] = &\
\color{#f00}{2^{3n - 1} + \half\,\delta_{n0}}
\end{align}