Combinatorial Proof Of Binomial Double Counting Let $a$, $b$, $c$ and $n$ be non-negative integers. 
By counting the number of committees 
consisting of $n$ sentient beings that can
be chosen from a pool of $a$ kittens, $b$
crocodiles and $c$ emus in two different
ways, prove the identity
$$\sum\limits_{\substack{i,j,k \ge 0; \\ i+j+k = n}} {{a \choose i}\cdot{b \choose j}\cdot{c \choose k} = {a+b+c \choose n}}$$
where the sum is over all non-negative integers $i$, $j$ and $k$ such that $i+j+k=n.$
I know that this is some kind of combinatorial proof. My biggest problem is that I've never really done a proof.

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$\ds{\sum_{i\ +\ j\ +\ k\ =\ n \atop{\vphantom{\LARGE A}i,\ j,\ k\  \geq\ 0}}
{a \choose i}{b \choose j}{c \choose k} = {a + b + c \choose n}:\ {\large ?}}$

\begin{align}&\color{#66f}{\large%
\sum_{i\ +\ j\ +\ k\ =\ n \atop{\vphantom{\LARGE A}i,\ j,\ k\  \geq\ 0}}
{a \choose i}{b \choose j}{c \choose k}}
=\sum_{\ell_{a},\ \ell_{b},\ \ell_{c}\  \geq\ 0}{a \choose \ell_{a}}
{b \choose \ell_{b}}{c \choose \ell_{c}}
\delta_{\ell_{a}\ +\ \ell_{b}\ +\ \ell_{c},\ n}
\\[3mm]&=\sum_{\ell_{a},\ \ell_{b},\ \ell_{c}\  \geq\ 0}{a \choose \ell_{a}}
{b \choose \ell_{b}}{c \choose \ell_{c}}\oint_{\verts{z}\ =\ 1}
{1 \over z^{-\ell_{a}\ -\ \ell_{b}\ -\ \ell_{c}\ +\ n\ +\ 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z^{n\ +\ 1}}
\bracks{\sum_{\ell_{a}\ \geq\ 0}{a \choose \ell_{a}}z^{\ell_{a}}}
\bracks{\sum_{\ell_{b}\ \geq\ 0}{b \choose \ell_{b}}z^{\ell_{b}}}
\bracks{\sum_{\ell_{a}\ \geq\ 0}{c \choose \ell_{c}}z^{\ell_{c}}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\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}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{a + b + c} \over z^{n\ +\ 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\color{#66f}{\large{a + b + c \choose n}}
\end{align}

