5. For a given rational number, represent it as a reduced fraction. Then find the product of its numerator and denominator. How many rational numbers between 0 and 1 have a product of their numerator and denominator equal to 20! ?
5. 20! has 8 prime factors: $2, 3, 5, 7, 11, 13, 17, 19$. A prime factor appearing in the denominator cannot appear in the numerator (and vice versa). For each prime factor, it can be in the numerator or in the denominator, so there are $2^8 = 256$ choices. However, not all of them are less than 1. In fact, the numbers can be paired as reciprocals of each other. Therefore, only $\frac{256}{2} = 128$ satisfy the requirement.