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?

That's called "double convolution": we have
$$
\eqalign{
  & \sum\limits_{0\, \le \,k\, \le \,n\,} {\left( \matrix{
  a + k \cr 
  k \cr}  \right)\left( \matrix{
  b - k \cr 
  n - k \cr}  \right)} \quad \left| \matrix{
  \,a,b \in \mathbb C \hfill \cr 
  \;0 \le n \in \mathbb Z \hfill \cr}  \right.\quad  =  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)\,} {\left( \matrix{
  a + k \cr 
  k \cr}  \right)\left( \matrix{
  b - k \cr 
  n - k \cr}  \right)}  =  \quad \quad (1)  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)\,} {\left( { - 1} \right)^{\,k} \left( \matrix{
   - a - 1 \cr 
  k \cr}  \right)\left( { - 1} \right)^{\,n - k} \left( \matrix{
  n - b - 1 \cr 
  n - k \cr}  \right)}  =   \quad \quad (2) \cr 
  &  = \left( { - 1} \right)^{\,n} \left( \matrix{
  n - a - b - 2 \cr 
  n \cr}  \right) =  \quad \quad (3)  \cr 
  &  = \left( \matrix{
  a + b + 1 \cr 
  n \cr}  \right)   \quad \quad (4) \cr} 
$$
where:
 (1) we can omit the sum bounds because they are implicit in the binomials, this simplifies the algebraic manouvres;
 (2) upper negation;
 (3) Vandermonde convolution;
 (4) upper negation again.
