Any closed formula for this summation involving binomial coefficients?

Question

I am stuck with writing a closed formula for the following summation: $$\sum_{i=0}^{n-m} (-1)^i {m \choose i} {n-m \choose i}^2$$

I would appreciate any help.

Answer 1

As I wrote in a comment, $$\sum_{i=0}^{n-m} (-1)^i {m \choose i} {n-m \choose i}^2=\, _3F_2(-m,m-n,m-n;1,1;1)$$ where appears the generalized hypergeometric function which is not a closed form. I do not see any way to simplify it.

If you fix $m$, you will get a polynomial in $n$ of degree $2m$ $$\left( \begin{array}{cc} m & \, _3F_2(-m,m-n,m-n;1,1;1) \\ 0 & 1 \\ 1 & -n^2 +2n\\ 2 & \frac{n^4}{4}-\frac{5 n^3}{2}+\frac{29 n^2}{4}-7 n+2 \\ 3 & -\frac{n^6}{36}+\frac{2 n^5}{3}-\frac{211 n^4}{36}+\frac{145 n^3}{6}-\frac{893 n^2}{18}+\frac{146 n}{3}-18 \\ 4 & \frac{n^8}{576}-\frac{11 n^7}{144}+\frac{389 n^6}{288}-\frac{113 n^5}{9}+\frac{38785 n^4}{576}-\frac{30869 n^3}{144}+\frac{6365 n^2}{16}-\frac{791 n}{2}+162 \end{array} \right)$$