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Consider the number 48.
There are five pairs of integers $a$ and $b$ ($a \leq b$) 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 $a \times b=n$, ($a \leq b$) 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 $C(100!)$

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