Interesting Binomial identity

Question

Fix $n,\ell, b \in \mathbb{N}$ with $2b\le n.$ Set $r=\min\{n-2b,\ell\}.$ Then $$\binom{n}{\ell} \sum_{i=0}^{r}(-1)^{i}\binom{n-i}{2b}\binom{\ell}{i} = {\binom{n}{2b}}\sum_{i=0}^{\ell}\binom{b}{i}\binom{b}{\ell-i}$$ The two formulas are (different) solutions of the same combinatorial problem. Although, I am wondering how we can prove this analytically.

Answer 1

The left sum reduces to $\binom{n-\ell}{2b-\ell}$, and the right sum reduces (via Vandermonde's identity) to $\binom{2b}{\ell}$. So the LHS and RHS are equal because of the known identity $$\binom{n}{k}\binom{n-k}{m-k}=\binom{n}{m}\binom{m}{k}.$$

Here's a generating function approach for the left sum: \begin{align} \sum_n \left(\sum_{i=0}^n (-1)^i \binom{\ell}{i} \binom{n-i}{2b}\right)z^n &=\left(\sum_n (-1)^n \binom{\ell}{n}z^n\right)\left(\sum_n \binom{n}{2b}z^n\right)\\ &=(1-z)^\ell \frac{z^{2b}}{(1-z)^{2b+1}} \\ &=\frac{z^{2b}}{(1-z)^{2b-\ell+1}} \\ &=\sum_n \binom{n-\ell}{2b-\ell}z^n \end{align}