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!