Combinatorial identity for $\sum_{k = 0}^n \binom{n - k + a}{n - k} \binom{k + b}{k}$ I am interested in
\begin{equation}
\sum_{k = 0}^n \binom{n - k + a}{n - k} \binom{k + b}{k} \tag{1},
\end{equation}
for non-negative integers $a$ and $b$. Mathematica tells me that
\begin{equation}
(1) = \binom{n + a + b + 1}{n}.
\end{equation} 
Is this a well-known identity and does it, for example, show up in any of the volumes of the Tables of Combinatorial Identities? If not, is there a combinatorial interpretation of this identity?
 A: Well, $(1)$ is just the coefficient of $x^n$ in the product between
$$ \sum_{m\geq 0}\binom{m+a}{m}x^m\quad\text{and}\quad \sum_{m\geq 0}\binom{m+b}{m}x^m, $$
hence, by stars and bars, the coefficient of $x^n$ the product between
$$ \frac{1}{(1-x)^{a+1}}\quad\text{and}\quad \frac{1}{(1-x)^{b+1}}, $$
hence $\color{red}{\binom{n+a+b+1}{n}}$ as claimed.
A: Yes, it is well known, as it is the Vandermonde convolution in a disguised form
$$
\eqalign{
  & \sum\limits_{k\, = \,0}^n {\left( \matrix{
  n - k + a \cr 
  n - k \cr}  \right)\left( \matrix{
  k + b \cr 
  k \cr}  \right)}  =   \cr 
  &  = \sum\limits_{k\, = \,0}^n {\left( { - 1} \right)^{\,n - k} \left( \matrix{
   - a - 1 \cr 
  n - k \cr}  \right)\left( { - 1} \right)^{\,k} \left( \matrix{
   - b - 1 \cr 
  k \cr}  \right)}  = ({\rm uppernegation})  \cr 
  &  = \left( { - 1} \right)^{\,n} \sum\limits_{k\, = \,0}^n {\left( \matrix{
   - a - 1 \cr 
  n - k \cr}  \right)\left( \matrix{
   - b - 1 \cr 
  k \cr}  \right)}  =   \cr 
  &  = \left( { - 1} \right)^{\,n} \left( \matrix{
   - a - b - 2 \cr 
  n \cr}  \right) = ({\rm Vandermonde}\;{\rm conv}{\rm .})  \cr 
  &  = \left( \matrix{
  n + a + b + 1 \cr 
  n \cr}  \right)({\rm uppernegation}) \cr} 
$$
With the "extended definition" of the binomial 
$$
\left( \matrix{
  x \cr 
  q \cr}  \right) = x^{\,\underline {\,q\,} } /q!\;{\rm iff}\;0 \le {\rm integer}\,q,\;0\;{\rm otherwise}
$$
the result is valid also for $a,b$ real or even complex.
