Question

Prove that the product of 2n consecutive negative integers is divisible by (2n)!

Answer 1

Let $2n$ negative integers be $\left(-r\right),\left(-r-1\right),\left(-r-2\right),....,...,\left(-r-2n+1\right)$.

Then, product = ${\left(-1\right)}^{2n}\left(r\right)\left(r+1\right)\left(r+2\right),....,...\left(r+2n-1\right)$
$$\begin{gathered} =\frac{\left(r-1\right)!\left(r\right)\left(r+1\right)\left(r+2\right)......\left(r+2n-1\right)}{\left(r-1\right)!} \\ =\frac{\left(r+2n-1\right)!}{\left(r-1\right)!} \\ =\frac{\left(r+2n-1\right)!}{\left(r-1\right)!\left(2n\right)!}\times \left(2n\right)! \\ ={}^{r+2n-1}{C}_{2n}\times \left(2n\right)! \end{gathered}$$
This is divisible by $\left(2n\right)!.$