$\newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}}
\newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{{\large{j,\ k,\ \ell\ \geq\ 0\ \atop j\ +\ k\ +\ \ell\ =\ n}}}
{a \choose j}{b \choose k}{c \choose \ell} & =
\sum_{j,\ k,\ \ell\ \geq\ 0}
{a \choose j}{b \choose k}{c \choose \ell}\
\overbrace{\oint_{\verts{z}\ =\ 1^{-}}
{1 \over z^{n + 1 - j - k - \ell}}\,{\dd z \over 2\pi\ic}}
^{\ds{\bracks{j + k + \ell = n}}}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{n + 1}}
\bracks{\sum_{j = 0}^{\infty}{a \choose j}z^{j}}
\bracks{\sum_{k = 0}^{\infty}{b \choose k}z^{k}}
\bracks{\sum_{\ell = 0}^{\infty}{c \choose \ell}z^{\ell}}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{1 \over z^{n + 1}}
\pars{1 + z}^{a}\,\pars{1 + z}^{b}\,\pars{1 + z}^{c}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{a + b + c} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic} =
\bbox[8px,border:0.1em groove navy]{a + b + c \choose n}
\end{align}
La última expresión se encuentra en expansión, en potencias de $\ds{z}$,
$\ds{\pars{1 + z}^{a + b + c}}$. La única contribución a la
la integral proviene de la '$\ds{z^{n}\!}$plazo' de
$\ds{\pars{1 + z}^{a + b + c}}$. Es decir,
$$
\bracks{z^{n}}\pars{1 + z}^{a + b + c} = {a + b + c \elegir n}
$$