Let a rational number $\frac{a}{b}$ in its lowest form where $a$,$b$ are integers, with $0 < \frac{a}{b}< 1$, b > 1. How many of these have $ab = 15!$

Question

Consider a rational number $\frac{a}{b}$ in its lowest form where $a$, $b$ are integers, with $0 < \frac{a}{b}< 1$, b > 1. How many of these have $ab = 15!$

Solution Given in Book:

$15!=2^{11}\cdot 3^6\cdot 5^3\cdot 7^2\cdot 11\cdot13$

Number of ways = Number of ways $15!$ can be expressed as product of $2$ relatively prime divisors = $2^{6–1} = 32$

Doubt: How do they arrived at the formula $2^{n-1}$.

Any Help is appreciated.

Answer 1

For each factor $2^{11}, 3^6, 5^3, 7^2, 11, 13$ choose whether to put it in the numerator or in the denominator. That gives you $2^6$ possible choices for $\frac{a}{b}$. But exactly half of them are bigger than $1$ (since $\frac{a}{b}>1$ if and only if $\frac{b}{a}<1$), so you need to divide by $2$.