Combinatorics That Looks Similar to Vandermonde's Identity How do I simplify:
$$\sum_{r = 0}^{\lfloor \frac{k}{2} \rfloor} \dbinom{k}{2r} \cdot \dbinom{n-k}{k-2r}?$$
Basically, the sum is: $\dbinom{k}{0} \cdot \dbinom{n-k}{k} + \dbinom{k}{2} \cdot \dbinom{n-k}{k-2} + \dbinom{k}{4} \cdot \dbinom{n-k}{k-4} \cdots.$
This seems very similar to Vandermonde's Identity but because we skip two everytime instead of one, I don't see how I can apply it to this problem.

Note that
$$\sum_{r\ge 0} a_{2r} = \sum_{r\ge 0} \frac{1+(-1)^r}{2}a_r.$$
Taking $$a_r=\binom{k}{r}\binom{n-k}{k-r}$$ yields
\begin{align}
\sum_{r\ge0} \binom{k}{2r}\binom{n-k}{k-2r} 
&= \sum_{r\ge0} \frac{1+(-1)^r}{2} \binom{k}{r}\binom{n-k}{k-r} \\
&= \frac{1}{2}\sum_{r\ge0} \binom{k}{r}\binom{n-k}{k-r} + \frac{1}{2}\sum_{r\ge0} (-1)^r\binom{k}{r}\binom{n-k}{k-r} \\
&= \frac{1}{2}\binom{n}{k} + \frac{1}{2}\sum_{r\ge0} (-1)^r\binom{k}{r}\binom{n-k}{k-r}.
\end{align}
I don't know whether the remaining sum can be simplified.
