# Combinatorial view of binomial identity

For the identity of binomials below, we have a combinatorial view.

$$\sum_{k=0}^n \binom{n}{k} k(n-k)=\binom{n}{2}2^{n-1}$$ For the left-hand side: Choose $$k$$ balls from n balls and paint them in red, then paint rest $$(n-k)$$ balls in blue. Finally, choose one red ball and one blue ball.

For the right-hand side: Choose $$2$$ balls from $$n$$ balls and paint one ball in red and the other in blue. And paint rest $$n-2$$ balls freely in red and blue.

Then, can you find the combinatorial view of the identity below? (I found this formula counting the number of a path in the grid. link).

$$\sum_{i=0}^{\min(x,y)}\binom{x}{i}\binom{y}{i}2^{x+y-(i+1)}=\sum_{i=0}^{\min(x,y)}2^{x+y-(i+1)} (-1)^i \frac{(x+y-i)!}{(x-i)!(y-i)!i!}$$

This seems to have something to do with the inclusion-exclusion principle. However, I cannot come up with a combinatorial view.

Left hand side count the way to put $$i$$ blocks of $$\rightarrow\ldots\rightarrow\uparrow\ldots\uparrow$$.
Right hand side count the way to put $$i$$ blocks of $$\rightarrow\ldots\rightarrow\uparrow\ldots\uparrow$$ incoherently in the block. Then arranging them using inclusion-exclusion principle.