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 $r \geq n$.

Answer 1

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?