Combinatorial Identity with Binomial Coefficients: $ {{a+b+c-1}\choose c} = \sum_{i+j=c} {{a+i-1}\choose i}{{b+j-1}\choose j} $ I got the following identity from commutative algebra. 
I am curious to see elegant elementary methods. 

$$ {{a+b+c-1}\choose c} = \sum_{i+j=c} {{a+i-1}\choose i}{{b+j-1}\choose j} $$

 A: It is just a case of Vandermonde's identity. 
The LHS is the coefficient of $x^c$ in $\frac{1}{(1-x)^{a+b}}$, by stars and bars. 
On the other hand, $\frac{1}{(1-x)^{a+b}}=\frac{1}{(1-x)^{a}}\cdot\frac{1}{(1-x)^b}$, hence:
$$\begin{eqnarray*} \binom{a+b+c-1}{c}=[x^c]\frac{1}{(1-x)^{a+b}}&=&\sum_{d=0}^{c}[x^d]\frac{1}{(1-x)^a}\cdot [x^{c-d}]\frac{1}{(1-x)^b}\\ &=& \sum_{i+j=c}\binom{a+i-1}{i}\binom{b+j-1}{j}, \end{eqnarray*}$$
as wanted.
A: For variety's sake here is a slightly different approach.
Suppose we seek to evaluate
$$\sum_{k=0}^c {k+a-1\choose a-1} {b-1+c-k\choose b-1}.$$
Introduce
$${b-1+c-k\choose b-1} = {b-1+c-k\choose c-k} = 
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c-k+1}} (1+z)^{b-1+c-k} \; dz.$$
Observe  that this  is zero  when $k\gt  c$ so  we may  extend  $k$ to
infinity to get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{b-1+c} 
\sum_{k\ge 0} {k+a-1\choose a-1} \frac{z^k}{(1+z)^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{b-1+c} 
\frac{1}{(1-z/(1+z))^a}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c+1}} (1+z)^{a+b-1+c} 
\frac{1}{(1+z-z)^a}
\; dz
\\ = {a+b+c-1\choose c}.$$
We can take $\epsilon \lt 1/2$ for the series to converge.
A: Using the identity
$$
\sum_{j=0}^n\binom{n-j}{k}\binom{j}{m}=\binom{n+1}{k+m+1}
$$
proven in this answer, we get
$$
\begin{align}
\sum_{i+j=c}\binom{a+i-1}{i}\binom{b+j-1}{j}
&=\sum_{i+j=c}\binom{a+i-1}{a-1}\binom{b+j-1}{b-1}\\
&=\binom{a+b+c-1}{a+b-1}\\
&=\binom{a+b+c-1}{c}\\
\end{align}
$$
