How many positive integers divide $20!$ $20!$ Has nos that are multiples of $2,3,4$ and so on. However, the total number of integers is large. So, please help me.
 A: Since $20!=2^{18}\cdot3^{8}\cdot5^{4}\cdot7^{2}\cdot11^{1}\cdot13^{1}\cdot17^{1}\cdot19^{1}$:


*

*$2$ can appear in every divisor between $0$ and $18$ times, i.e., $19$ combinations

*$3$ can appear in every divisor between $0$ and $8$ times, i.e., $9$ combinations

*$5$ can appear in every divisor between $0$ and $4$ times, i.e., $5$ combinations

*$7$ can appear in every divisor between $0$ and $2$ times, i.e., $3$ combinations

*$11$ can appear in every divisor between $0$ and $1$ times, i.e., $2$ combinations

*$13$ can appear in every divisor between $0$ and $1$ times, i.e., $2$ combinations

*$17$ can appear in every divisor between $0$ and $1$ times, i.e., $2$ combinations

*$19$ can appear in every divisor between $0$ and $1$ times, i.e., $2$ combinations


Therefore, the number of divisors of $20!$ is $19\cdot9\cdot5\cdot3\cdot2\cdot2\cdot2\cdot2=41040$.
A: Hint:
Use Legendre's formula:
For each prime $p\le n$,  the exponent of $p$ in the prime decomposition of $n!$ is
$$v_p(n!)=\biggl\lfloor\frac{n}{p}\biggr\rfloor+\biggl\lfloor\frac{n}{p^2}\biggr\rfloor+\biggl\lfloor\frac{n}{p^3}\biggr\rfloor+\dotsm$$
The number of prime divisors of $n!$ is then
$$\prod_{\substack{ p\;\text{prime}\\p\le n}}\bigl(v_p(n!)+1\bigr).$$
