I was working on proving the following fact of probability. Suppose $A_1,\dots,A_n$ are events, and define \begin{align*} S_1 &= \sum_{1\le i \le n}P(A_i) \\ S_2 &= \sum_{1\le i < j \le n}P(A_i\cap A_j) \\ S_3 &= \sum_{1\le i < j < k \le n}P(A_i\cap A_j\cap A_k) \\ \vdots & \qquad\qquad\vdots \end{align*} and so on. Then the probability of exactly $m$ events occurring is $$ p(m) = S_m - {m+1\choose m}S_{m+1}+{m+2\choose m}S_{m+2}-\dots+(-1)^{n-m}{n\choose m}S_n. $$ As a part of proving this identity, I ended up proving the following result about binomial coefficients.
Let $1\le m \le n$ and $1\le r\le n-m$. Then $$ \color{blue}{{m+r\choose m} = \sum_{j=1}^{r}(-1)^{j+1}{m+j\choose m}{m+r\choose m+j}}. $$ I suspect there has to be a nice combinatorial interpretation of this alternating sum, and I am hoping someone would make it clear to me if it exists.
Just to be clear, I am not asking for a proof of the identity in blue, which I already have. I am asking if there is a nice way to see the result by counting. After all, the left-hand side is the number of ways to choose $m$ objects from $m+r$ objects, and the right-hand side is some kind of inclusion-exclusion-y thing, also involving binomial coefficients. Thanks!
As a bonus, there is a similar alternating sum on this site, and I expect that these two are related. If there is a connection, $+\epsilon$ credit for explaining the connection between these two sums.