Proving $\sum \limits_{j=0}^{n+1- (a+b)} \binom{a+j-1}{a -1} \binom{n- a-j}{b -1} =\binom{n}{a+b-1} $ I have run into quiete a tricky sum, I am reasonably sure I know what is sums to but I have been unable to prove it
$$
\forall a,b,n \in \mathbb{N}, \ n\geq a+b \geq 2:\quad
\sum \limits_{j=0}^{n+1- (a+b)}
  \binom{a+j-1}{a -1}
  \binom{n- a-j}{b -1}
=\binom{n}{a+b-1}
$$
I have tried to evaluate it for quiete a number of particular values and it seems to hold. I've tried to prove it by induction however that didn't seem to lead anywhere but I could be wrong.

We seek to verify that with $n\ge a+b\ge 2$
$$\sum_{j=0}^{n+1-a-b} {a+j-1\choose a-1} {n-a-j\choose b-1}
= {n\choose a+b-1}.$$
The LHS is
$$\sum_{j=0}^{n+1-a-b} {a+j-1\choose a-1}
{n-a-j\choose n+1-a-b-j}
\\ = [z^{n+1-a-b}] (1+z)^{n-a}
\sum_{j\ge 0} {a+j-1\choose a-1} \frac{z^j}{(1+z)^j} $$
Here we have extended to infinity because the coefficient extractor
enforces the upper limit. Continuing,
$$[z^{n+1-a-b}] (1+z)^{n-a}
\frac{1}{(1-z/(1+z))^a}
\\ = [z^{n+1-a-b}] (1+z)^{n}
= {n\choose n+1-a-b} = {n\choose a+b-1}.$$
This is the claim.
