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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.