Yet another sum involving binomial coefficients.

Question

Given $A,B,N \in \mathbb N$ Is there a closed form for this expression?

$$\sum_{n=1}^N n \binom{A}n \binom{B}{N-n} $$

If there is such, can you give a proof?

EDIT: $A,B \geq N$

Answer 1

For the full solution of what @Yuval Filmus is saying,

\begin{align} \sum_{n=1}^{N}{n\binom{A}{n}\binom{B}{N-n}} &= \sum_{n=1}^{N}{A\binom{A-1}{n-1}\binom{B}{N-n}}\\ &= A\sum_{n=1}^{N}{\binom{A-1}{n-1}\binom{B}{N-n}}\\ &= A\binom{A+B-1}{N-1}, \end{align} Where the last equality comes from http://en.wikipedia.org/wiki/Vandermonde's_identity.

Answer 2

Hint: $n\binom{A}{n} = A\binom{A-1}{n-1}$.

Answer 3

From what @Yuval_Filmus says

\begin{align} \sum_{n=1}^{N}{n\binom{A}{n}\binom{B}{N-n}} & = A\sum \binom{A-1}{n-1}\binom{B}{N-n}\\ &= A\sum \left( \binom{A}n - \binom{A-1}{n} \right)\binom{B}{N-n} \\ &= A\left( \binom{A+B}{N} - \binom{A+B-1}{N}\right)\\ &= \frac {AN}{A+B}\binom{A+B}{N} \end{align}

The third equality comes from Vandermonde's identity.