# Difference between revisions of "2019 AMC 10A Problems/Problem 25"

The following problem is from both the 2019 AMC 10A #25 and 2019 AMC 12A #24, so both problems redirect to this page.

## Problem

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

$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

## Solution

The main insight is that

$$\frac{(n^2)!}{(n!)^{n+1}}$$

is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$. Thus,

$$\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}$$

is an integer if $n^2 \mid n!$, or in other words, if $n \mid (n-1)!$. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem. There are $15$ primes between $1$ and $50$, inclusive, so there are 15 + 1 = 16 terms for which

$$\frac{(n^2-1)!}{(n!)^{n}}$$

is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of 2 in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$, as there are more factors of p in the denominator than the numerator. Thus all 16 values of n make the expression not an integer and the answer is $50-16=\boxed{\mathbf{(D)}\ 34}$.