Rationals from 25!

• Observation 1: $25! = 2^{22} \cdot 3^{10} \cdot 5^6 \cdot 7^3 \cdot 11^2 \cdot 13 \cdot 17 \cdot 19 \cdot 23$
• Observation 2: $m$ and $n$ are comprised of and only of these factors. Also, if a prime factor divides $m$ , it cannot divide $n$ . Hence, the factors must occur in clusters, in which all primes are together.

The problem thus reduces to partitioning the 9 clusters $2^{22}, 3^{10} , 5^6 , 7^3 , 11^2 , 13 , 17 , 19 , 23$ into $m$ and $n$ .

For any cluster, there are two choices, it could either be in $m$ or in $n$ . So, there are $2^9$ ways to split them.

How do we eliminate the cases where $m > n$ ? Well, for any pair $(m, n)$ , if $m > n$ , switching them obtains a pair $(n, m)$ , where $n < m$ . Hence there are half as many ways to split the numbers in the way we desire.

That'd be $\frac{2^9}{2} = \boxed{2^8}$ ways.