How it can be shown that:
$$\sum_{k=0}^{2r}\left(-1\right)^{k}\binom{n}{k}\binom{n}{2r-k}=\left(-1\right)^{r}\binom{n}{r}$$
$$ \begin{align} \sum_{k=0}^{2r}\left(-1\right)^{k}\binom{n}{k}\binom{n}{2r-k} &=\left(-1\right)^{n}\sum_{k=0}^{2r}\binom{n}{k}\binom{k-2r-1}{n-2r+k}\tag{1}\\ &=\left(-1\right)^{n}\sum_{k=0}^{2r}\binom{n}{k}\sum_{l}^{}\binom{k}{n-l}\binom{-2r-1}{-2r+k+l}\tag{2}\\ &=\left(-1\right)^{n}\sum_{k=0}^{2r}\sum_{l}^{}\binom{n}{n-l}\binom{l}{n-k}\binom{-2r-1}{-2r+k+l}\tag{3}\\ &=\left(-1\right)^{n}\sum_{l}^{}\binom{n}{l}\binom{l-2r-1}{l-2r+n}\tag{4}\\ &=\sum_{l}^{}\binom{n}{l}\binom{n}{2r-l}\left(-1\right)^{l}\tag{5} \end{align} $$
- $(1)$: Pascal's rule and negative binomial coefficients
- $(2)$: Converse of Vandermonde's convolution
- $(3)$: applying the identity $\binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{n-k}$
- $(4)$: Vandermonde's convolution
- $(5)$: negative binomial coefficients
The final answer depends on the alternating sign Vandermonde convolution.
It's known that:
$$\sum_{k=0}^{r}\binom{n}{k}\binom{n}{r-k}\left(-1\right)^{k}=\left(-1\right)^{\frac{r}{2}}\binom{n}{\frac{r}{2}}\tag{I}$$
For $r$ even.
So setting $2r \mapsto r$ follows the result, but how even $\text{(I)}$ can be proved?
Source : math.wvu.edu
