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

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 $n$?

Answer 1

The Laguerre Polynomials are defined as $$L_n(x)=\sum_{j=0}^{n} (-1)^{j} \frac{{n \choose j}}{j!} x^J,~~~(1)$$ these are well known to follow the orthogonality condition as $$\int_{0}^{\infty} e^{-x} L_m(x)~ L_n(x)~ dx=\delta_{m,n}~,~~~~(2)$$ where $\delta_{m,n}$ is the Kroneckor delta function, If we insert (1) in (2), we get $$\int_{0}^{\infty}\sum_{i=0}^{m} \sum_{0}^{n} (-1)^{i+j} \frac{{m \choose i}}{i!} \frac{{n \choose j}}{j!} x^{i+j} e^{-x}~ dx= \delta_{m,n}.$$ Finally, $\int_{0}^{\infty} x^k e^{-x}~dx= k!$, leads to $$\sum_{i=0}^{m} \sum_{j=0}^{n} (-1)^{i+j} {m \choose i} {n \choose j} {i+j \choose i} = \delta_{m,n}.$$ Hence the result.