Is there a simpler way to compute this sum? For any given positive integers $m$, $n$ and $q$, such that $m\leq n$ the following sum
$$S_p= \sum_{p=0}^m(-1)^{p+q} \binom mp \binom mq \binom np \binom nq\frac{p! q! (m+n-p-q)!}{m! n!}$$
is equal to $1$ if $q=0$ and $0$ otherwise.
The result can be obtained remarking that $S_p$ of the form 
$$ \sum_{p=0}^m(-1)^p\binom mp P(m-p)\tag{1}$$ where $P$ is a polynomial of degree $m-q$. One then uses the following property of the form (1) for polynomials $P$ of degree at most $m$, that the sum is equal to $m! a_m$, where $a_m$ is the coefficient of $X^m$ in $P(X)$. 


*

*What is the name of this property (or a reference, I only know it from wikipedia) ? 

*Is there a simpler way to demonstrate the result without using this particular property ?

 A: We need to calculate the sum
$$S = \sum_{p \ge 0}t_p = \sum_{p \ge 0}(-1)^{p+q} \binom mp \binom mq \binom np \binom nq\frac{p! q! (m+n-p-q)!}{m! n!}$$
where $t_p = 0$ for $p > m$ because $\binom{m}{p}  = 0$ for $p > m$.
The term ratio $$\frac{t_{p+1}}{t_p} = \frac{(-1)(m-p)(n-p)(p+1)}{(p+1)(p+1)(m+n-p-q)} = \frac{(p-m)(p-n)}{(p+q-m-n)(p+1)}$$
is a rational function of the summation index $p$.
Thus $S$ is a hypergeometric series given by
$$\begin{align}S &= {_2}F_1\left[\begin{matrix} -m & -n\\q-m-n\end{matrix} \text{ ; } 1\right] \cdot t_0\end{align}$$
which fits the Gauss hypergeometric identity to give
$$S = \frac{\Gamma(q)\Gamma(q-m-n)}{\Gamma(q-m)\Gamma(q-n)} \cdot t_0$$
We know that
$\begin{align}t_0 &=(-1)^q \binom{m}{q}\binom{n}{q}\frac{q!(m+n-q)!}{m!n!}\\&=(-1)^q\frac{\Gamma(m+n-q+1)}{\Gamma(q+1)\Gamma(m-q+1)\Gamma(n-q+1)}\end{align}$
after writing the factorials and binomial coefficients in terms of the Gamma function.
Substituting this into $S$ gives
$\begin{align}S &= \color{darkblue}{(-1)^q} \frac{\Gamma(q)}{\color{darkblue}{\Gamma(q+1)}}\frac{\Gamma(q-m-n)\color{darkblue}{\Gamma(m+n-q+1)}}{\color{darkgreen}{(}\Gamma(q-m)\color{darkblue}{\Gamma(m-q+1)}\color{darkgreen}{)}\cdot \color{darkgreen}{(}\Gamma(q-n)\color{darkblue}{\Gamma(n-q+1)}\color{darkgreen}{)}}\end{align}$
Pairs of terms find themselves amenable to the Euler reflection formula, giving an expression in terms of the sine function.
$\begin{align}S &= (-1)^{q} \frac{1}{q}\frac{\frac{\pi}{\sin{\pi(q-m-n)}}}{\frac{\pi}{\sin{\pi(q-m)}}\frac{\pi}{\sin{\pi(q-n)}}} \\\\ &= \frac{(-1)^{q}}{q\pi }\frac{\sin{(q\pi - m\pi)}\sin{(q\pi - n\pi)}}{\sin{(q\pi - (m+n)\pi )}} \end{align}$
The trigonometric identities for symmetry and periodicity reduce the sum to
$\begin{align}S &= \frac{(-1)^{q}}{q\pi}\frac{(-1)^{m}\sin{( q\pi)}(-1)^n\sin{(q \pi)}}{(-1)^{m+n}\sin{(q\pi)}}\\ &= \frac{(-1)^{q}\sin{( q\pi)}}{q\pi} \\ &= \begin{cases} 0 & q \ne 0 \\ 1 & q = 0 \end{cases}\end{align}$
In terms of the Kronecker delta, the sum becomes  $$\color{darkred}{S = \delta_q}$$
Note: I referenced chapter 3 - The Hypergeometric Database of the wonderful book A=B for this problem.
A: I really do not know how much this could help : using a CAS, what I found is that $$S=\sum_{p=0}^m(-1)^{p+q} \binom mp \binom mq \binom np \binom nq\frac{p! q! (m+n-p-q)!}{m! n!}=\frac{(-1)^q \sin (\pi  q)}{\pi  q}$$
