ordered pair of $(p,q)$ in $20!$ 
A rational number given in the form
$\displaystyle \frac{p}{q},\;\;p,q\in \mathbb{Z}^{+}\;,\frac{p}{q}\in(0,1)$ and $p,q$ are coprime to each other.If $pq=20!.$ Then number of ordered pair of $p,q$ are

what i try
$20!=2^{18}\cdot 3^{8}\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$
Let $p=2^{a_{1}}\cdot 3^{a_{2}}\cdot 5^{a_{3}}\cdot 7^{a_{4}}\cdot 11^{a_{5}}\cdot 13^{a_{6}}\cdot 17^{a_{7}}\cdot 19^{a_{8}}$
$q=2^{b_{1}}\cdot 3^{b_{2}}\cdot 5^{b_{3}}\cdot 7^{b_{4}}\cdot 11^{b_{5}}\cdot 13^{b_{6}}\cdot 17^{b_{7}}\cdot 19^{b_{8}}$
$0\leq a_{1}\leq 18,0\leq a_{2}\leq 8,0\leq a_{3}\leq 4,0\leq a_{4}\leq 2,0\leq a_{5}\leq 1$
$0\leq a_{6}\leq 1,0\leq a_{7}\leq 1,0\leq a_{8}\leq 1$
and $0\leq b_{1}\leq 18,0\leq b_{2}\leq 8,0\leq b_{3}\leq 4,0\leq b_{4}\leq 2,0\leq b_{5}\leq 1$
$0\leq b_{6}\leq 1,0\leq b_{7}\leq 1,0\leq b_{8}\leq 1$
How do i solve it Help me please
 A: Two notes to help you continue:


*

*You didn't use the fact that $p,q$ are coprime. That fact allows you to remove many options, since you cannot have $p=2^k\cdots$ and $q=2^l\cdots$. Either one or the other must be $2^{18}\cdots$, while the other must be odd.

*First, count the total number of pairs $(p, q)$ such that $pq=20!$. Then, for each pair $(p,q)$, think about the two numbers $x=\frac pq$ and $y=\frac qp$. Can they both be greater than $1$? Can they both be smaller than $1$?

A: That $\gcd(p,q)=1$ means each prime power factor of $20!$ must either be wholly in $p$ or wholly in $q$. The fraction $\frac pq$ can never equal $1$, and if it is greater than $1$ then $\frac qp$ is less than $1$ and vice versa.
There are $2^8$ ways to assign the prime powers to either $p$ or $q$, so the number of fractions less than $1$ is half that, or $128$.
A: Since $p.q=20!$ hence I can treat this like there are eight distinct objects i.e $2^{18},3^8,5^4,7^2,11,13,17,19$ need to be distributed into two different groups $p , q$ where 
$$ $$
Total number of ways are $2^8$ since each object can go in any one of the group. And since $20!$ is not a perfect square hence $p\neq q$ .
$$ $$ 
For $\frac{p}{q} \in (0,1)$ we need to consider unordered pairs of solutions i.e $\frac{2^8}{2}=2^7=128$
