I am trying to simplify the following summation of products of binomial coefficients: $$\sum_{k=0}^n \binom{k+b}{a} \binom{y+n-k}{y}$$ where $b > a$ (specifically, $b = 2a+1$).
I have searched through some of the usual resources (Gradshteyn and Ryzhik, HW Gould, the DLMF, Abramowtiz and Stegun, Wolfram's online resources) and have come up empty so far.
If anyone has any ideas on how to approach this problem, I would be very thankful!
You can show that $$ \sum_{k=a-b}^n \binom{k+b}{a} \binom{y+n-k}{y} %=\sum_{h=0}^{n+b-a}\binom{h+a}a\binom{y+n+b-a-h}{y} =\binom{b+n+y+1}{a+y+1}.\tag{*} $$ That is, your summation is a nice one minus some missing terms. Therefore, your summation is $$ \boxed{\binom{b+n+y+1}{a+y+1}-\sum_{k=a-b}^{-1}\binom{k+b}a\binom{y+n-k}y}$$ To prove $(*)$, use $$ \sum_{i=k}^{n-h} \binom{i}k\binom{n-i}h=\binom{n+1}{k+h+1}\tag{**}$$ and reindex appropriately. A combinatorial proof of this $(**)$: to choose a subset of $\{1,2,\dots,n,n+1\}$ of size $k+h+1$ whose $(k+1)^{st}$ smallest element is equal to $i+1$, you choose the $k$ elements below $i+1$, and the $h$ elements which are above $i+1$.