Show that $\binom{n-1}{r-1}=\sum^{r-1}_{k=0}(-1)^k\binom rk\binom{n+r-k-1}{r-k-1}$

Question

This is an identity that I couldn't prove.

$$\binom{n-1}{r-1}=\sum^{r-1}_{k=0}(-1)^k\binom rk\binom{n+r-k-1}{r-k-1}.$$

I believe that the identity is provable through combinatorics, but I can't prove it.

Answer 1

Consider the sum $$a_1+a_2+\cdots+a_r=n$$

The number of positive solutions (strictly greater than $0$) is ${n-1 \choose r-1}$ as a solution to the sum is the same as placing $r-1$ sticks in the $n-1$ spaces between $n$ balls to generate $a_1,a_2,\ldots,a_r$ balls separated by sticks.

Fact $1$: the number of positive solutions is also

(The number of non-negative solutions) $-$ (The number of non-negative solutions where at least one of the $a_i$'s is zero) $+$ (The number of non-negative solutions where at least two of the $a_i$'s are zero) $-$ (The number of non-negative solutions where at least three of the $a_i$'s are zero) $+\cdots$ and so on by the inclusion-exclusion principle.

Lemma $1$: the number of non-negative solutions for $a_1+a_2+\cdots+a_r=n$ where at least $k$ of the $a_i$'s are zero is $${r \choose k}{n+r-k-1 \choose r-k-1}$$

Proof:

First we choose the $k$ terms to be zero, hence the $r\choose k$ part, and we are left with $r-k$ non-negative terms adding up to $n$. Rewrite them as

$$b_1 + b_2+\cdots+b_{r-k}=n\tag1$$

where each $b_j$ represents a not-chosen-to-be-zero $a_i$ and by definition each $b_j$ is non-negative and can be zero as well.

Now we rewrite the sum as the following

$$(b_1+1)+(b_2+1)+\cdots+(b_{r-k}+1)=n+r-k\tag2$$

The number of solutions in $(1)$ and $(2)$ are exactly the same and $(2)$ only involves positive (strictly greater than $0$) terms so the number of solutions is

$$n+r-k-1\choose r-k-1$$.

By fact $1$ and lemma $1$, the result is therefore proven.