Prove that $\sum_{i=0}^{k}\binom{k-i}{b}\binom{i}{a-1}=\binom{k+1}{a+b}$ I am trying to prove this identity from an exercise:
Given $k$, $a$, $b$, prove that
$$
\sum_{i=0}^{k}\binom{k-i}{b}\binom{i}{a-1}=\binom{k+1}{a+b}
$$
However, I'm having trouble using existing identities to prove this. Any help would be much appreciated.

Start from
$$\sum_{q=a-1}^{k-b} {k-q\choose b} {q\choose a-1}
= \sum_{q=0}^{k+1-b-a} {k+1-a-q\choose b} {q+a-1\choose a-1}
\\ = \sum_{q=0}^{k+1-a-b}
{k+1-a-q\choose k+1-a-b-q} {q+a-1\choose a-1}
\\ = [z^{k+1-a-b}] (1+z)^{k+1-a} \sum_{q=0}^{k+1-a-b}
{q+a-1\choose a-1} \frac{z^q}{(1+z)^q}.$$
The coefficient extractor enforces the range and we get
$$[z^{k+1-a-b}] (1+z)^{k+1-a} \sum_{q\ge 0}
{q+a-1\choose a-1} \frac{z^q}{(1+z)^q}
\\ = [z^{k+1-a-b}] (1+z)^{k+1-a}
\frac{1}{(1-z/(1+z))^a}
\\ = [z^{k+1-a-b}] (1+z)^{k+1-a}
\frac{(1+z)^a}{(1+z-z)^a}
\\ = [z^{k+1-a-b}] (1+z)^{k+1}
= {k+1\choose k+1-a-b} = {k+1\choose a+b}.$$
This  is  the  claim. BTW  when  $k-q\lt  b$  or  $k-b\lt q$  we  have
$(k-q)^\underline{b} = 0.$ (This zero value does not include $q=k$ and
$b=0$ because we required $k-q\lt b$.)  Similarly when $0\le q\lt a-1$
we have  $q^\underline{a-1} =  0.$ (This zero  value does  not include
$q=0$ and  $a=1$ because  we required $q\lt  a-1$.)  Applies  to $k,b$
non-negative integers  and $a$ a  positive integer. For the  sum range
not to be empty we also need $k-b\ge a-1$ or $k+1\ge a+b.$ 
