Consider the number $48$.
There are five pairs of integers $a$ and $b$ ($a\le b$) such that $a\times b=48$: $(1,48)$, $(2,24)$, $(3,16)$, $(4,12)$ and $(6,8)$.
It can be seen that both $6$ and $8$ have $4$ divisors.
So of those five pairs one consists of two integers with the same number of divisors.

In general:
Let $C(n)$ be the number of pairs of positive integers $a\times b=n$, ($a\le b$) such that $a$ and $b$ have the same number of divisors;
so $C(48)=1$.

You are given $C(10!)=3$: $(1680,2160)$, $(1800,2016)$ and $(1890,1920)$.

Find $C(100!)$.