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.
 A: On the righthand side, we form a committee of size $n$ by lumping all the animals together (so there are $a + b + c$ animals total) and choosing $n$ of them.
On the lefthand side, we form a committee of size $n$ by first choosing exactly $i$ kittens, $j$ crocodiles, and $k$ emus (subject to $i + j + k = n$). For fixed $i$, $j$, and $k$, there are 
$$
\binom{a}{i}\binom{b}{j}\binom{c}{k}
$$
such committees. Summing over all possible partitions of the integer $n$ means we've formed the committees in every possible way, which is precisely what is counted on the righthand side.
A: Assuming we want to choose from $a+b$ first before choosing  from $c$ such that $n_a+n_b=r$ then the number of ways of choosing is ($p$ from $a$ and $r-p$ from $b$) $$\sum_{p=0}^r \binom ap\displaystyle\binom b{r-p}$$
Afterward $c$ can must fill $n-(n_a+n_b)=n-r$ spaces so we choose $n-r$ from $c$ and $r$ from $a+b$ which implies that one can rewrite the sum as $$\sum\limits_{\substack{i,j,k \ge 0;\\ i+j+k = n}} {a \choose i}\cdot{b \choose j}\cdot{c \choose k} = \sum_{r=0}^n \binom c{n-r}\left( \sum_{p=0}^r \binom ap\binom b{r-p}\right)=\underbrace{\sum_{r=0}^n \sum_{p=0}^r \binom c{n-r} \binom ap\binom b{r-p}}$$
Note that if $r>n$ then $\displaystyle \binom nr =0$, so the above expression is valid.
Product of three polynomials $$\begin{array}{ll}\sum^a_{i=0}a_ix^i\cdot \sum^b_{j=0}b_jx^j\cdot\sum^c_{k=0}c_kx^k &= \sum^{a+b}_{m=0}\left( \sum_{p=0}^m a_pb_{m-p}\right)x^m \cdot\sum^c_{k=0}c_kx^k \quad\text {we use }d_m=\sum_{p=0}^m a_pb_{m-p}\\&=\sum^{a+b}_{m=0}d_mx^m \cdot\sum^c_{k=0}c_kx^k \\&= \sum^{a+b+c}_{n=0}\left( \sum_{r=0}^n d_rc_{n-r}\right)x^n =\sum^{a+b+c}_{n=0}\left( \sum_{r=0}^n c_{n-r}\left(\sum_{p=0}^r a_pb_{r-p}\right)\right)x^n\\&=\sum^{a+b+c}_{n=0} \sum_{r=0}^n  \sum_{p=0}^ra_p b_{r-p}c_{n-r}\cdot x^n \quad \quad \quad (1) \end{array}$$
By the binomial theorem $(1+x)^{a+b+c}$ $$ \begin{array}{ll}= \sum^{a+b+c}_{n=0}\binom {a+b+c}{n}x^n\\ = (1-x)^a(1-x)^b(1-x)^c\\=\left(\sum^a_{i=0}\binom ai x^i \right)\left(\sum^b_{j=0}\binom bj x^j \right)\left(\sum^c_{k=0}\binom ck x^k \right) \\=\sum^{a+b+c}_{n=0} \underbrace{\sum_{r=0}^n \sum_{p=0}^r \binom c{n-r} \binom ap\binom b{r-p}} x^n  \quad \text{after using result (1)} \end{array}$$
Observe that the under-braced expression is exactly the one under-braced above and it is the coefficient of of $x^n$,  we can then equate it to the coefficient binomial expansion i;e;
$$\binom {a+b+c}{n}=\sum_{r=0}^n \sum_{p=0}^r \binom c{n-r} \binom ap\binom b{r-p}$$
This is the Vandermonde's identity for three polynomials and it is the only combinatorial proof I know of.  Thanks to Douglas S. Stones who pointed this out to me in a question I asked.
One can do the same for $4$, $5$, ... but it is exhausting so proof by induction or the one given by Austin is sufficient.
A: $\newcommand{\+}{^{\dagger}}
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$\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}

