Question

For how many integers $n$ between $1$ and $50,$ inclusive, is $\dfrac{(n^2-1)!}{(n!)^n}$ an integer? (Recall that $0! = 1.$ )

$31$

$32$

$33$

$34$

$35$

One fact that greatly helps with this problem is realizing that $\dfrac{(n^2)!}{(n!)^{n + 1}}$ is always an integer.

This is because it is the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n.$

Now, we get that $\dfrac{(n^2 - 1)!}{(n!)^n} = \dfrac{(n^2)!}{(n!)^{n + 1}} \cdot \dfrac{n!}{n^2}.$

Therefore, we need to find when $n^2 \div n!,$ or when $n \div (n - 1)!.$

This condition is false if $n = 4,$ or if $n$ is prime. $n = 4$ is too large for it to divide $3!.$

$n$ cannot be prime because $(n - 1)!$ does not contain any numbers where $n$ could be a factor.

If $n$ is not $4$ and not prime, this works since $n$ can be decomposed into $2$ numbers both less than $n$ that are found in $(n - 1)!.$

There are $15$ primes less than $50,$ and adding on $4,$ we get that there are $16$ values for $n$ that do not work.

Therefore, the desired answer is $50 - 16 = 34.$

Thus, D is the correct answer.