How to find$\sum_{i,j,k\in \mathbb{Z}}\binom{n}{i+j}\binom{n}{j+k}\binom{n}{i+k}$ for $n \in \mathbb{N}$ 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.
 A: The key is given any $(u,v,w) \in \mathbb{Z}^3$, one can find
$(i,j,k) \in \mathbb{Z}^3$ such that
$$\begin{cases}
u &= i + j,\\
v &= j + k,\\
w &= k + i
\end{cases}$$
when and only when $u + v + w$ is even. Furthermore, the $(i,j,k)$ associated to $(u,,v,w)$ is unique if it exists. This leads to
$$
 \sum_{i,j,k \in \mathbb{Z}}\binom{n}{i+j}
\binom{n}{j+k}\binom{n}{k+i}
= \sum_{\substack{u,v,w \in \mathbb{Z},\\ u+v+w\text{ even}}}
\binom{n}{u}\binom{n}{v}\binom{n}{w}
= \sum_{\substack{0 \le u, v, w \le n\\ u+v+w\text{ even}}}
\binom{n}{u}\binom{n}{v}\binom{n}{w}\\
= \frac12 \sum_{0 \le u, v, w \le n}\left(1 + (-1)^{u+v+w}\right)\binom{n}{u}\binom{n}{v}\binom{n}{w}
= \frac12\left\{\left[\sum_{u=0}^n \binom{n}{u}\right]^3
+ \left[\sum_{u=0}^n (-1)^u\binom{n}{u}\right]^3
\right\}\\
= \frac12\left[(1+1)^{3n} + (1-1)^{3n}\right]
\stackrel{\color{blue}{[1]}}{=} \frac12 \left[2^{3n} + \begin{cases}0,& n > 0\\1,&n = 0\end{cases}\right]
= \begin{cases}2^{3n-1},&n > 0\\1, &n = 0\end{cases}
$$
Notes


*

*$\color{blue}{[1]}$ - thanks for @FelixMartin pointing out the special case at $n = 0$.

A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\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}
