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.
 A: 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.
A: 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
$\{ a+j+1,\dots,n\}$.
A: 
Consider $2$ points on Cartesian plane, $O(0,0)$ and $P(a+b-1,n+1-a-b)$ . Number of shortest routes from $O(0,0)$ to $P(a+b-1,n+1-a-b)$ is $$\binom{(n+1-a-b) + (a+b-1)}{a+b-1} = \binom{n}{a+b-1}$$ We will count this another way.
Consider points $A_0(a-1,0) , A_1(a-1,1) \ldots A_j(a-1,j) \ldots A_i(a-1,n+1-a-b)$ and $B_1(a,0), B_2(a,2) \ldots B_j(a,j) \ldots B_i(a,n+1-a-b)$ . Now join the points $A_1$ to $B_1$, $A_2$ to $B_2$ , $\dots$ $A_j$ to $B_j$ $\ldots$ and lastly $A_i$ to $B_i$ .Every shortest route from $O(0,0)$ to $P(a+b-1,n+1-a-b)$ must go through any of these segments . And thus number of shortest routes from $O(0,0)$ to $P(a+b-1,n+1-a-b)$ that go through $A_jB_j$ is $$\binom{(a-1)+(j)}{a-1}\binom{(a+b-1 - a) + (n+1-a-b - j)}{(a+b-1) - (a)} = \binom{a+j-1}{a-1}\binom{n-a-j}{b-1}$$ .
And thus the result follows by addition principle.
