Prove that $\sum_{k=1}^a (-1)^{a-k} {a \choose k} {b+k \choose b+1} = {b \choose a-1}$ (Feller Volume 1, 2.12.12) Prove that  for integers $0 < a <b$
$$\sum_{k=1}^a (-1)^{a-k} {a \choose k} {b+k \choose b+1} = {b \choose a-1}.$$
Hint: Using (12.4) show that (12.11) is a special case of (12.9). 
(12.4) For any $a>0$, 
$${-a \choose k} = (-1)^k {a+k-1 \choose k}.$$ 
(12.9) For any positive integers $a, b, n,$
$$\sum_{v=0}^n {a \choose v}{b \choose n-v} = {a+b \choose n}.$$ 
(12.11) 
$$\sum_{v=0}^n {n \choose v}^2 = {2n \choose n}.$$
I can show that (12.11) is a special case of (12.9) by letting $a = b= n$. I am not sure why the author suggests us to use (12.4). Also, how does this help us to prove the original question? The right side of (12.4) looks similar with $(-1)^{a-k}{a \choose k}$ in the summation, so I replace $k$ by $a-k$ in the right side of (12.4), but it doesn't look helpful. I really appreciate if you give some help. 

An elementary one:
$$\sum_{k=1}^a (-1)^{a-k} {a \choose k} {b+k \choose b+1}=\sum_{k=1}^a(-1)^{a-k}{a \choose k} {b+k \choose k-1}$$
$$=\sum_{k=1}^a(-1)^{a-1}{a \choose k} {-b-2 \choose k-1}=(-1)^{a-1}\sum_{k=1}^a{a \choose a-k} {-b-2 \choose k-1}$$
Now use Vandermonde's identity to get:
$$=(-1)^{a-1}\binom{a-b-2}{a-1}=\binom{b}{a-1}$$
Which is the claim.
