Proving a Vandermonde-like combinatorial identity

Question

I want to prove the identity $\sum\limits _{r=0}^{n} {{p+r} \choose {r}}\cdot {{q+n-r} \choose {n-r}}={{n+p+q+1} \choose n}$. I think induction does not work because replacing $n$ by $(n+1)$ we have $LHS=\sum\limits _{r=0}^{n+1} {{p+r} \choose r}\cdot {{q+(n+1)-r} \choose {(n+1)-r}}={p \choose 0}\cdot {{q+n+1} \choose {n+1}}+{{p+1} \choose 1}\cdot {{q+n} \choose n}+{{p+2} \choose 2}\cdot {{q+n-1} \choose {n-1}}+\cdots {{p+n+1} \choose {n+1}}\cdot {q \choose 0}$, where the second term in the product shifts by one place.

Answer 1

To expand on my comment:

By Vandermonde, we have $$\sum_{r = 0}^n \binom P r \binom Q{n - r} = \binom {P + Q}n$$ for all integers $P, Q \geq n$. We can view both sides as polynomials in $P, Q$ with rational coefficients. Since the equality holds for all integers $P, Q \geq n$, we see that it is in fact an equality of polynomials.

This allows us to set $P = -p - 1$ and $Q = -q - 1$ in the identity. We have $$\binom {-p-1}r = \frac{(-p-1)\cdots(-p-r)}{1\cdots r} = (-1)^r \frac{(p+r) \cdots(p+1)}{1\cdots r} = (-1)^r \binom{p + r}r$$ and similarly $$\binom{-q-1}{n-r} = (-1)^{n-r}\binom{q+n-r}{n-r}, \binom{-p-q-2}n = (-1)^n\binom{p+q+n+1}n$$ from which the original equality follows.