Is there a way to express $(n-i)!(n-j)!(2i)!(2j)!$ in terms of $n$ and $r=i+j$?

Question

I have been attempting to simplify the double sum: $$\sum_{i=0}^n \sum_{j=0}^n \frac{(-1)^{i+j} (2i+2j)!}{(n-i)!(n-j)!(2i)!(2j)!2^{i+j}(i+j)!}$$ And so what I am attempting to do is rewrite it in terms of $r=i+j$, and we can simplify the previous expression into: $$\sum_{i=0}^n \sum_{j=0}^n \frac{(-1)^{r} (2r)!}{(n-i)!(n-j)!(2i)!(2j)!2^{r}r!}$$ And so now I cannot figure out how to put $(n-i)!(n-j)!(2i)!(2j)!$ in terms of $r$ and $n$. Does anyone know of a way this could be done?

Answer 1

This is not possible because the expression does not only depend on $r$ and $n$. For example, take $n=5$, $i=2$, $j=2$, to get $6\cdot 6\cdot 24\cdot 24=20736$, then take $i=1$, $j=3$ to get $24\cdot 2\cdot 2\cdot 720=69120$ from the same expression, even though $n$ and $r$ are the same.