Binomial coefficient identity with alternating summands

Question

Is there a nice closed form for

$\sum_{k=0}^{l} (-1)^k \binom{n}{k} \binom{n-k}{l-k}.$

I feel like there must be but I cannot find it.

Answer 1

Start by proving that: $$\binom{n}{k} \binom{n-k}{l-k} = \binom{n}{l}\binom{l}{k}$$

Then,

\begin{align} \sum_{k=0}^{l} (-1)^k \binom{n}{k} \binom{n-k}{l-k} &= \sum_{k=0}^{l} (-1)^k \binom{n}{l}\binom{l}{k} \\ &= \binom{n}{l} \left(1 - 1\right)^l = 0 \end{align}