TITLE: Binomial identity
QUESTION [8 upvotes]: I'd like to get a hint to prove the following identity:
$$\tag{1}\sum_{\nu}(-1)^{\nu}\displaystyle \binom{a}{\nu}\binom{n-\nu}{r}=\binom{n-a}{n-r} .$$
The original statement reads "By specialization derive from (12.9) the identity..." where (12.9) refers to:
$$\tag{2}\sum_{\nu}\binom{a}{\nu}\binom{b}{n-\nu}=\binom{a+b}{n}. $$
The problem suggests using $$\tag{3} \binom{-a}{k} = (-1)^k\binom{a+k-1}{k}.$$
I assumed that "by specialization" means "start with (2) and derive $(1)$", so I changed the sign of $a$ in (2), then I used $(3)$, but it didn't work or I don't know how to proceed from there, so I'd appreciate any help.
Thanks!

REPLY [5 votes]: As long as both $n-\nu$ and $r$ are non-negative integers, we have
$$
\begin{align}
(-1)^\nu\binom{n-\nu}{r}
&=(-1)^\nu\binom{n-\nu}{n-\nu-r}\\
&=(-1)^{n-r}\binom{-r-1}{n-\nu-r}\tag{1}
\end{align}
$$
Using $(1)$, we get
$$
\begin{align}
\sum_{\nu}(-1)^{\nu}\binom{a}{\nu}\binom{n-\nu}{r}
&=(-1)^{n-r}\sum_{\nu}\binom{a}{\nu}\binom{-r-1}{n-\nu-r}\\
&=(-1)^{n-r}\binom{a-r-1}{n-r}\\
&=\binom{n-a}{n-r}\tag{2}
\end{align}
$$