# 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 ?

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.

• Thank you very much for this answer. I definitely learned something reading it. And thank you also for the reference to the book A=B. – Tom-Tom Apr 22 '14 at 19:44

I really do not know how much this could help : using a CAS, what I found is that $$S=\sum_{p=0}^m(-1)^{p+q} \binom mp \binom mq \binom np \binom nq\frac{p! q! (m+n-p-q)!}{m! n!}=\frac{(-1)^q \sin (\pi q)}{\pi q}$$

• So the question is how did it get to this expression... – Tom-Tom Apr 17 '14 at 10:15
• I really do not know either ! I was just fascinated by the expression; so I tried. Sorry for not helping you more with this problem. The result is quite amazing. Cheers. – Claude Leibovici Apr 17 '14 at 10:18
• The result on the polynomial concerning the expression (1) is proved in the following paper: An Algebraic Identity Leading to Wilson's Theorem Sebastián Martín Ruiz The Mathematical Gazette Vol. 80, No. 489 (Nov., 1996) (pp. 579-582). It is absolutely not easy. I doubt there is another demonstration of the result skipping this part. But after your comment, I think there cold be an hypergeometric identity hidden behind it (the way your CAS has followed, probably). – Tom-Tom Apr 17 '14 at 11:57