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.

If you  plan to select a set of $a+b-1$ items from $\{1,\ldots,n\}$,
then the $a$'th item must have the form $a+j$ for some
$j \in \{0,\dots,  n+1 -a-b\}$.
Once you fix $j$, you still need to select
$a-1$ items   from $\{1,\ldots,a+j-1\}$, and $b-1$ items from
$\{ a+j+1,\dots,n\}$.
