Proving the binomial identity $\sum_{j=0}^k {n \choose j}\cdot {m \choose k-j}={n+m \choose k}$ when $n,m$ are negative integers What I'm really interested in is the identity:
$$\displaystyle \sum_{j=0}^k {j+r-1 \choose j}\cdot {k-j +s-1 \choose k-j}={k+r+s-1 \choose k}$$
It arises in the proof of the convolution formula of two negative binomial Random Variables, for example.
What I know is the identity:$\displaystyle { j+r-1 \choose j} = {-r \choose j} (-1)^j $
Hence what I need to show is $$ \displaystyle \sum_{j=0}^k {-r \choose j} \cdot {-s \choose k-j}={-r-s \choose k}  $$ aka the  binomial convolution formula for $m = -r, n = -s$ when $s$ and $r$ are positive intergers.
How do I do that?
 A: For instance, you can use induction on $n$ to show that if $p_{n}(x)$ is the polynomial:
$$p_{0}(x)=1$$
$$p_{n}(x)=(x)_n=x(x-1)\cdots(x-n+1),\ \text{for each}\ n\geq 1$$
then
$$p_{n}(x+y)=\sum_{k=0}^{n}{n\choose k}p_{k}(x)p_{n-k}(y)$$

Your identity is equivalent to the one above.

Recall that if $n\in\mathbb{N}$ and $x$ is a variable you can write:
$${x\choose n}=\frac{(x)_{n}}{n!}$$
The falling factorial is a family of polynomials of binomial type like the fmailies $f_{n}(x)=x^n$, $g_{n}(x)=x(x+1)\cdots(x+n-1)$, Abel polynomials, Touchard polynomials.
A: Here is an answer based upon generating functions. It is convenient  to use the coefficient of operator  $[z^n]$ to  denote the coefficient of  $z^n$  in  a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{j=0}^k}&\color{blue}{\binom{j+r-1}{j}\binom{k-j+s-1}{k-j}}\\
&=\sum_{=0}^k\binom{-r}{j}(-1)^j\binom{-s}{k-j}(-1)^{k-j}\\
&=(-1)^k\sum_{j=0}^\infty[z^j](1+z)^{-r}[u^{k-j}](1+u)^{-s}\tag{1}\\
&=(-1)^k[u^k](1+u)^{-s}\sum_{j=0}^\infty u^j[z^j](1+z)^{-r}\tag{2}\\
&=(-1)^k[u^k](1+u)^{-s}(1+u)^{-r}\tag{3}\\
&=(-1)^k[u^k](1+u)^{-r-s}\\
&=(-1)^k\binom{-r-s}{k}\\
&\color{blue}{=\binom{k+r+s-1}{k}}
\end{align*}
  and the Chu-Vandermonde identity follows.

Comment:


*

*In (1) we apply the coefficient of operator twice and set the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.

*In (2) we use the linearity of the coefficient of operator and apply the rule $[z^{p}]z^qA(z)=[z^{p-q}]A(z)$.

*In (3) we apply the substitution rule of the coefficient of operator with $u=z$
\begin{align*}
A(u)=\sum_{j=0}^\infty a_j u^j=\sum_{j=0}^\infty u^j [z^j]A(z)
\end{align*}
