There is a well-known formula for the product of the binomial coefficients: $$\binom{n}{a}\binom{n}{b}=\sum_{i=0}^{min(a,b)}\binom{a+b-i}{i,a-i,b-i}\binom{n}{a+b-i}$$
I'm interested in a different formula for the expansion of the product, where the RHS only contains binomial coefficients of the form $\binom{n+i}{a+b}$: $$\binom{n}{a}\binom{n}{b}=\sum_{i=0}^{a+b}\binom{a}{i}\binom{b}{i}\binom{n+i}{a+b}$$
More generally, $$\binom{n+c}{a}\binom{n+d}{b}=\sum_{i=0}^{a+b}\binom{a-c+d}{i-c}\binom{b-d+c}{i-d}\binom{n+i}{a+b}$$
This holds whenever $0\le c \le a$ and $0 \le d \le b$, and it comes from this MO answer regarding a quantized version of the formula.
I'm almost certain that I'm not the first to find this formula, but identities involving binomial coefficients are hard to sort through. I was wondering if someone remembers having seen this before and could point me to a reference.
We can prove the identity while we wait for a reference to appear. We seek with $0\le c\le a$ and $0\le d\le b$
$${n+c\choose a} {n+d\choose b} = \sum_{q=0}^{a+b} {a-c+d\choose q-c} {b-d+c\choose q-d} {n+q\choose a+b}.$$
We get for the RHS
$$[z^{a+b}] \frac{1}{1-z} \sum_{q\ge 0} z^q {a-c+d\choose a+d-q} {b-d+c\choose b+c-q} {n+q\choose a+b} \\ = [z^{a+b}] \frac{1}{1-z} [w^{a+d}] (1+w)^{a-c+d} [v^{b+c}] (1+v)^{b-d+c} \sum_{q\ge 0} z^q w^q v^q {n+q\choose a+b} \\ = [z^{a+b}] \frac{1}{1-z} [w^{a+d}] (1+w)^{a-c+d} [v^{b+c}] (1+v)^{b-d+c} [u^{a+b}] (1+u)^n \\ \times \sum_{q\ge 0} z^q w^q v^q (1+u)^q \\ = [z^{a+b}] \frac{1}{1-z} [w^{a+d}] (1+w)^{a-c+d} [v^{b+c}] (1+v)^{b-d+c} [u^{a+b}] (1+u)^n \frac{1}{1-zwv(1+u)} \\ = - [z^{a+b+1}] \frac{1}{1-z} [w^{a+d+1}] (1+w)^{a-c+d} [v^{b+c}] (1+v)^{b-d+c} \\ \times [u^{a+b}] (1+u)^{n-1} \frac{1}{v-1/z/w/(1+u)}.$$
The contribution from $v$ is
$$\;\underset{v}{\mathrm{res}}\; \frac{1}{v^{b+c+1}} (1+v)^{b-d+c} \frac{1}{v-1/z/w/(1+u)}.$$
Here we see that the residue at infinity is zero, so we may use minus the residue at $v=1/z/w/(1+u)$ (residues sum to zero):
$$- (1+u)^{b+c+1} z^{b+c+1} w^{b+c+1} \frac{(zw(1+u)+1)^{b-d+c}}{z^{b-d+c} w^{b-d+c} (1+u)^{b-d+c}} \\ = - (1+u)^{d+1} z^{d+1} w^{d+1} (zw(1+u)+1)^{b-d+c}.$$
Substitute into the remaining extractors,
$$[z^{a+b-d}] \frac{1}{1-z} [w^a] (1+w)^{a-c+d} [u^{a+b}] (1+u)^{n+d} (zw(1+u)+1)^{b-d+c}.$$
The contribution from $z$ is
$$\;\underset{z}{\mathrm{res}}\; \frac{1}{z^{a+b-d+1}} \frac{1}{1-z} (zw(1+u)+1)^{b-d+c}.$$
Using the the inequalities as stated in the preliminaries we find again that the residue at infinity is zero, and we may use minus the residue at $z=1.$ We also note that there is no pole at $z=-1/w/(1+u).$ Collecting the remaining two extractors, we get
$$[w^a] (1+w)^{a-c+d} [u^{a+b}] (1+u)^{n+d} (w(1+u)+1)^{b-d+c} \\ = [w^a] (1+w)^{a-c+d} [u^{a+b}] (1+u)^{n+d} (1+w+uw)^{b-d+c} \\ = [w^a] (1+w)^{a-c+d} [u^{a+b}] (1+u)^{n+d} \sum_{q=0}^{b-d+c} {b-d+c\choose q} (1+w)^{b-d+c-q} u^q w^q \\ = [u^{a+b}] (1+u)^{n+d} \sum_{q=0}^{b-d+c} {b-d+c\choose q} {a+b-q\choose a-q} u^q \\ = \sum_{q=0}^{b-d+c} {b-d+c\choose q} {a+b-q\choose a-q} {n+d\choose a+b-q}.$$
Here we have by construction that the rightmost two binomial coefficients are zero when the lower index goes negative. This means that we can replace the upper range by $a$. If $a\lt b-d+c$ we get for $q\gt a$ that the second binomial coefficient is zero. If $b-d+c\lt a$ we get for $q\gt b-d+c$ that the first binomial coefficient is zero. We thus have
$$\sum_{q=0}^a {b-d+c\choose q} {a+b-q\choose a-q} {n+d\choose a+b-q}.$$
Now observe that
$${a+b-q\choose a-q} {n+d\choose a+b-q} = \frac{(n+d)!}{(a-q)!\times b! \times (n+d+q-a-b)!} \\ = {n+d\choose b} {n+d-b\choose a-q}.$$
We thus obtain
$${n+d\choose b} \sum_{q=0}^a {b-d+c\choose q} {n+d-b\choose a-q}.$$
Here we may use Vandermonde which yields
$${n+d\choose b} {n+c\choose a}.$$
This is the claim. It is interesting to try this sum with the major CAS.