Given a [rational number](https://artofproblemsolving.com/wiki/index.php/Rational_number), write it as a [fraction](https://artofproblemsolving.com/wiki/index.php/Fraction) in lowest terms and calculate the product of the resulting [numerator](https://artofproblemsolving.com/wiki/index.php/Numerator) and [denominator](https://artofproblemsolving.com/wiki/index.php/Denominator). For how many rational numbers between $0$ and $1$ will $20_{}^{}!$ be the resulting [product](https://artofproblemsolving.com/wiki/index.php/Product)?
If the fraction is in the form $\frac{a}{b}$, then $a < b$ and $\gcd(a,b) = 1$. There are 8 [prime numbers](https://artofproblemsolving.com/wiki/index.php/Prime_number) less than 20 ($2, 3, 5, 7, 11, 13, 17, 19$), and each can only be a factor of one of $a$ or $b$. There are $2^8$ ways of selecting some [combination](https://artofproblemsolving.com/wiki/index.php/Combination) of numbers for $a$; however, since $a<b$, only half of them will be between $0 < \frac{a}{b} < 1$. Therefore, the solution is $\frac{2^8}{2} = \boxed{128}$.