TITLE: The double sum $\sum_{j=0}^{n} \sum_{i=0}^{m} (-1)^{i+j} {m \choose i} {n \choose j} {i+j \choose i} =0~ \mbox{or}~1$
QUESTION [8 upvotes]: For non-negative integers $m$ and $n$, the double sum $$\sum_{j=0}^{n} \sum_{i=0}^{m} (-1)^{i+j} {m \choose i} {n \choose j} {i+j \choose i}$$
can be checked to be 0 or 1. How can one show this by hand, along with the conditions on $m$ and $n$?

REPLY [2 votes]: This is a slight variation of @MarkoRiedels nice answer. We  use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n }{k}\tag{1}
\end{align*}

We obtain for integral $m,n\geq 0$:
  \begin{align*}
\color{blue}{\sum_{p=0}^n}&\color{blue}{\sum_{q=0}^m(-1)^{p+q}\binom{n}{p}\binom{m}{q}\binom{p+q}{q}}\\
&=\sum_{p=0}^n\binom{n}{p}(-1)^p\sum_{q=0}^m\binom{m}{q}(-1)^q[z^q](1+z)^{p+q}\tag{2}\\
&=[z^0]\sum_{p=0}^n\binom{n}{p}(-1)^p(1+z)^p\sum_{q=0}^m\binom{m}{q}\left(-\frac{1+z}{z}\right)^q\tag{3}\\
&=[z^0]\sum_{p=0}^n\binom{n}{p}(-1)^p(1+z)^p\left(1-\frac{1+z}{z}\right)^m\tag{4}\\
&=(-1)^m[z^m]\sum_{p=0}^n\binom{n}{p}(-1)^p(1+z)^p\tag{5}\\
&=(-1)^m[z^m]\left(1-(1+z)\right)^n\\
&=(-1)^{m+n}[z^m]z^n\\
&\,\,\color{blue}{=[[m=n]]}\tag{6}
\end{align*}
and the claim follows.

Comment:

In (2) we use the coefficient of operator according to (1).
In (3) we do some rearrangements and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
In (4) we apply the binomial theorem.
In (5) we do some simplifications and apply the rule from (3) again.
In (6) we use Iverson Brackets.