Proof: $\binom{x+y+n-1}{n} = \sum_{k=0}^{n} \binom{x+n-k-1}{n-k} \binom{y+k-1}{k}$ I wanted to prove following equation
$\binom{x+y+n-1}{n} = \sum_{k=0}^{n} \binom{x+n-k-1}{n-k} \binom{y+k-1}{k}$
Using Vandermonde's identity 
$\binom{a+b}{t} = \sum_{k=0}^{t} \binom{a}{t-k} \binom{b}{k}$
whereas $a=x+n-k-1$ and $b=y+k-1$ but it doesn't add up. 
Where is my mistake? How do i proof it?

We have for our sum
$$\sum_{k=0}^n {x+n-k-1\choose n-k} {y+k-1\choose k}
\\ = [z^n] (1+z)^{x+n-1}
\sum_{k=0}^n {y+k-1\choose k} z^k (1+z)^{-k}.$$
The coefficient extractor controls the range of the sum:
$$[z^n] (1+z)^{x+n-1}
\sum_{k\ge 0} {y+k-1\choose k} z^k (1+z)^{-k}
\\ = [z^n] (1+z)^{x+n-1}
\frac{1}{(1-z/(1+z))^y} 
\\ = [z^n] (1+z)^{x+n-1} (1+z)^y =
[z^n] (1+z)^{x+y+n-1} \\ = {x+y+n-1\choose n}$$
as claimed.
