imc

Note that for all positive $n$, we have $$\frac{n!(n+1)!}{2}$$ $$⟹\frac{(n!{)}^2\cdot (n+1)}{2}$$ $$⟹(n!{)}^2\cdot \frac{n+1}{2}$$ We must find a value of $n$ such that $(n!{)}^2\cdot \frac{n+1}{2}$ is a perfect square. Since $(n!{)}^2$ is a perfect square, we must also have $\frac{n+1}{2}$ be a perfect square. In order for $\frac{n+1}{2}$ to be a perfect square, $n+1$ must be twice a perfect square. From the answer choices, $n+1=18$ works, thus, $n=17$ and our desired answer is $\boxed{\frac{17!18!}{2}}$