Prove $\sum_{i=0}^{\min(x,y)}\binom{x}{i}\binom{y}{i}2^{x+y-(i+1)}=\sum_{i=0}^{\min(x,y)}2^{x+y-(i+1)} (-1)^i \frac{(x+y-i)!}{(x-i)!(y-i)!i!}$ I found the formula below counting the path on the grid (link).
$$\sum_{i=0}^{\min(x,y)}\binom{x}{i}\binom{y}{i}2^{x+y-(i+1)}=\sum_{i=0}^{\min(x,y)}2^{x+y-(i+1)} (-1)^i \frac{(x+y-i)!}{(x-i)!(y-i)!i!}$$
How to prove this (I prefer more intuitive way)? I check this formula by a program up to $x,y\leq10$ and this seems correct.
My idea:
Suppose that $x>y$.
There is a formula which looks similar to this question.
\begin{align}
&[t^y] (1+t)^x(1+t)^y \\
=&\sum_{i=0}^{\min(x,y)}\binom{x}{i}\binom{y}{i} \\
=&\binom{x+y}{y}
\end{align}
Note that $[t^y] f(t)$ corresponds to the coefficient of the term $t^y$ of $f(t)$. Let's change the coefficients to meet our formula.
\begin{align} 
 &[t^y] \frac{1}{2}(2+t)^x(2+2t)^y \\
=&\sum_{i=0}^{\min(x,y)}\binom{x}{i}\binom{y}{i}2^{x+y-(i+1)}
\end{align}
I expect that by counting the coefficient of $t^y$ nicely, we can get $\sum_{i=0}^{\min(x,y)}2^{x+y-(i+1)} (-1)^i \frac{(x+y-i)!}{(x-i)!(y-i)!i!}$.
 A: Using the symmetry in the problem  statement we may suppose that $x\le
y.$ With $m,n$ for $x,y$ we then seek to prove
$$\sum_{q=0}^m {m\choose q} {n\choose q} 2^{-q}
= \sum_{q=0}^m {m+n-q\choose m-q} {n\choose q} 2^{-q} (-1)^q.$$
Start with the RHS to get
$$\sum_{q=0}^m  {n\choose q} 2^{-q} (-1)^q
[z^{m-q}] (1+z)^{m+n-q}
\\ = [z^m] (1+z)^{m+n}
\sum_{q=0}^m  {n\choose q} 2^{-q} (-1)^q
z^q (1+z)^{-q}.$$
Now the coefficient  extractor $[z^m]$ combined with  the factor $z^q$
enforces the range and we may continue with
$$[z^m] (1+z)^{m+n}
\sum_{q\ge 0}  {n\choose q} 2^{-q} (-1)^q
z^q (1+z)^{-q}
\\ = [z^m] (1+z)^{m+n}
\left(1 - \frac{z}{2(1+z)}\right)^n
\\ = \frac{1}{2^n} [z^m] (1+z)^{m} (2+z)^n.$$
On the other hand we have for the LHS
$$\sum_{q=0}^m {m\choose m-q} {n\choose q} 2^{-q}
\\ = \sum_{q=0}^m {n\choose q} 2^{-q}
[z^{m-q}] (1+z)^m
\\ = [z^{m}] (1+z)^m
\sum_{q=0}^m {n\choose q} 2^{-q} z^q.$$
The coefficient extractor once more enforces the range:
$$[z^{m}] (1+z)^m
\sum_{q\ge 0} {n\choose q} 2^{-q} z^q
\\ = [z^{m}] (1+z)^m \left(1+ \frac{z}{2}\right)^n
\\ = \frac{1}{2^n} [z^{m}] (1+z)^m (2+z)^n.$$
The  LHS and  the  RHS have  an identical  constant  factor, the  same
coefficient extractor and  the same argument thereto, and  we have the
claim.
