Show that for $r,n \in \mathbb{N}$ with $r \geq n$, $\sum_{i=0}^{n-1} (-1)^i \binom{n}{i} \binom{r+n-i-1}{r} = \binom{r-1}{n-1}$

Question

Show that $$\sum_{i=0}^{n-1} (-1)^i \binom{n}{i} \binom{r+n-i-1}{r} = \binom{r-1}{n-1}$$ where $r,n \in \mathbb{N}$ with $r \geq n$.

Can you please edit your post in order to show us what you've been trying? Otherwise, no one will solve your exercise for you ;) — Olivier Roche, Nov 15 '19 at 07:47
Please read: https://math.stackexchange.com/help/how-to-ask ; And: There is no indication whether the question should be answered combinatorially or analytically. — user90369, Nov 15 '19 at 09:04
See also: Proving $\binom {n-1}{r-1}=\sum_{k=0}^r(-1)^k\binom r k \binom{n+r-k-1}{r-k-1}$ — Martin Sleziak, Aug 12 '21 at 07:56

score 0 · Answer 1 · answered Nov 15 '19 at 07:48

Hint: ${ r+k-1 \choose r} = {r +k-1 \choose k-1} $ is the number of ways to distribute $r$ identical things amongst $k$ people.

Hint: The $(-1)^n {n\choose i}$ is very reminiscent of Principle of Inclusion and Exclusion.

Hint: How many ways are there to distribute $r$ identical things amongst $k$ people, so that [CONDITION X] is equal to the LHS and RHS?

1 Answers1