TITLE: A binomial identity
QUESTION [15 upvotes]: I was wandering if someone knows an elementary proof of the following identity:
$$
\frac{(a)_n (b)_n}{(n!)^2} = \sum_{k=0}^n (-1)^k {1-a-b \choose k} 
\frac{(1-a)_{n-k}(1-b)_{n-k}}{((n-k)!)^2}\ ,
$$
where $a,b$ are arbitrary real numbers,
$$(x)_0=1,\quad 
(x)_n:=x(x+1)\cdots(x+n-1) \quad 
\mbox{for $n\geq 1$}
$$ 
is the Pochhammer's symbol, and 
$$
{x\choose 0}=1,\quad 
{x\choose k} = \frac{x(x-1)\cdots (x-k+1)}{k!} \quad 
\mbox{for $k\geq 1$}$$
is a binomial coefficient.
The proof that I know uses the Hypergeometric differential equation. One has to continue analytically the solutions along a path connecting two singular points. This could be done by some well known integral representation of the Hypergeometric function. 
I think that there should be a combinatorial proof. Since this is an identity between polynomials in $a$ and $b$, it is enough to prove it for $a$ and $b$ negative integers, i.e., we may assume that $a=-p$ and $b=-q$ where $p,q\in \mathbb{Z}_{\geq 0}$. The identity then turns into
$$
{p\choose n}{q\choose n} = 
\sum_{k=0}^n (-1)^k {1+p+q\choose k}{p+n-k\choose n-k}{q+n-k\choose n-k}.
$$
I did not try very hard to proof the above identity and I did not search the literature that much, but since it comes from an interesting subject I think that it is worth finding an alternative proof.

REPLY [2 votes]: Suppose we seek to evaluate
$$\sum_{k=0}^n (-1)^k
{1+p+q\choose k} {p+n-k\choose n-k} {q+n-k\choose n-k}$$
which is claimed to be
$${p\choose n}{q\choose n}.$$
Introduce
$${p+n-k\choose n-k}
= \frac{1}{2\pi i}
\int_{|z_1|=\epsilon} \frac{(1+z_1)^{p+n-k}}{z_1^{n-k+1}} 
\; dz_1$$
and
$${q+n-k\choose n-k}
= \frac{1}{2\pi i}
\int_{|z_2|=\epsilon} \frac{(1+z_2)^{q+n-k}}{z_2^{n-k+1}} 
\; dz_2.$$
Observe that  these integrals vanish when  $k\gt n$ and  we may extend
$k$ to infinity.
We thus obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon} 
\frac{(1+z_1)^{p+n}}{z_1^{n+1}} 
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon} 
\frac{(1+z_2)^{q+n}}{z_2^{n+1}} 
\\ \times \sum_{k\ge 0} {1+p+q\choose k} (-1)^k
\frac{z_1^k z_2^k}{(1+z_1)^k (1+z_2)^k}
\; dz_2\; dz_1.$$
This is
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon} 
\frac{(1+z_1)^{p+n}}{z_1^{n+1}} 
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon} 
\frac{(1+z_2)^{q+n}}{z_2^{n+1}} 
\\ \times
\left(1-\frac{z_1 z_2}{(1+z_1)(1+z_2)}\right)^{p+q+1}
\; dz_2\; dz_1$$
or
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon} 
\frac{(1+z_1)^{n-q-1}}{z_1^{n+1}} 
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon} 
\frac{(1+z_2)^{n-p-1}}{z_2^{n+1}} 
(1+ z_1 + z_2)^{p+q+1}
\; dz_2\; dz_1$$
Supposing that $p\ge n$ and $q\ge n$ this may be re-written as
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon} 
\frac{1}{z_1^{n+1} (1+z_1)^{q+1-n}} 
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon} 
\frac{1}{z_2^{n+1} (1+z_2)^{p+1-n}} 
\\ \times (1+ z_1 + z_2)^{p+q+1}
\; dz_2\; dz_1$$
Put $z_2 = (1+z_1) z_3$ so that $dz_2 = (1+z_1) \; dz_3$ to get
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon} 
\frac{1}{z_1^{n+1} (1+z_1)^{q+1-n}} 
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon} 
\frac{1}{(1+z_1)^{n+1} z_3^{n+1} (1+(1+z_1)z_3)^{p+1-n}} 
\\ \times (1+ z_1)^{p+q+1} (1+z_3)^{p+q+1}
\; (1+z_1) \; dz_3\; dz_1$$
which is
$$\frac{1}{2\pi i}
\int_{|z_1|=\epsilon} 
\frac{(1+z_1)^p}{z_1^{n+1}} 
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon} 
\frac{1}{z_3^{n+1} (1+ z_3 + z_1 z_3)^{p+1-n}} 
\\ \times (1+z_3)^{p+q+1}
\; dz_3\; dz_1
\\ = \frac{1}{2\pi i}
\int_{|z_1|=\epsilon} 
\frac{(1+z_1)^p}{z_1^{n+1}} 
\frac{1}{2\pi i}
\int_{|z_2|=\epsilon} 
\frac{(1+z_3)^{n+q}}{z_3^{n+1} (1 + z_1 z_3 /(1+z_3))^{p+1-n}} 
\; dz_3\; dz_1$$
Extracting the residue for $z_1$ first we obtain
$$\sum_{k=0}^n {p\choose n-k}
\frac{(1+z_3)^{n+q}}{z_3^{n+1}}
{k+p-n\choose k}
(-1)^k \frac{z_3^k}{(1+z_3)^k}.$$
The residue for $z_3$ then yields
$$\sum_{k=0}^n (-1)^k {p\choose n-k}
{k+p-n\choose k} {n-k+q\choose n-k}.$$
The sum term here is
$$\frac{p!\times (p+k-n)!\times (q+n-k)!}
{(n-k)! (p+k-n)! \times k! (p-n)! \times (n-k)! q!}$$
which simplifies to
$$\frac{p!\times n! \times (q+n-k)!}
{(n-k)! \times n!\times  k! (p-n)! \times (n-k)! q!}$$
which is
$${n\choose k} {p\choose n}{q+n-k\choose q}$$
so we have for the sum
$${p\choose n}
\sum_{k=0}^n {n\choose k} (-1)^k {q+n-k\choose q}.$$
To evaluae the remaining sum we introduce
$${q+n-k\choose q}
= \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{q+n-k}}{v^{q+1}} \; dv$$
getting for the sum
$${p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q+n}}{v^{q+1}}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+v)^k} \; dv
\\ = {p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q+n}}{v^{q+1}}
\left(1-\frac{1}{1+v}\right)^n \; dv
\\ = {p\choose n}
\frac{1}{2\pi i}
\int_{|v|=\epsilon}
\frac{(1+v)^{q}}{v^{q-n+1}} \; dv
= {p\choose n} {q\choose q-n}$$
which is $${p\choose n} {q\choose n}.$$
This concludes the argument.