A rational number is given in the form: A rational number is given in the form: $\frac{p}{q}$; $\frac{p}{q} \in (0,1)$ and $p$ and $q$ are coprime to each other. If $pq= 10!$ , what would be the number of ordered pairs $(p,q)$?

A beginning step would be noticing that $10!$ consist of $4$ primes $(2, 3, 5, 7)$. But how do I proceed? Taking combinations of two numbers from these $4$ primes doesn't seem right and making cases of this also doesn't yield the right result.
 A: I believe the answer is 8, but let me go through my reasoning. Since the numbers are coprime, you can't give one prime to both $p$ and $q$, as it seems you have realized. So, we are really counting the number of ways of assigning the primes $\{2,3,5,7\}$ to $p$ and $q$. This is equivalent to picking a subset of these numbers and assigning them to $p$ (the rest need to go to $q$). Thus, this would give us $2^4 = 16$ options. However, given such a pair $(p, q)$, exactly one of $q/p$ and $p/q$ is going to be in the desired range $(0, 1)$. Thus, we cut our options in half and we get 8 such ordered pairs $(p, q)$.
A: Figure out not just the prime factors of $10!$ but the powers as well...i.e. the complete prime factorization of $10!$.
$10!$ is a big (relatively speaking) number, but it's not a monster.  We can express it as $1*2*3*4*5*6*7*8*9*10 = 1*2*3*2^2*5*(2*3)*7*2^3*3^2*(2*5) = 2^8\cdot 3^4\cdot 5^2\cdot 7$
So if $\gcd(p,q) = 1$ and $pq = 2^8\cdot 3^4\cdot 5^2\cdot 7$ then
.... can you take it from here?...

Warning:  The following is free thinking so it will include indirect inefficient reasoning.  The point is to replicate how one would think about solving it; not haw to solve it in hindsight efficiently

$p = 2^a3^b5^c7^d$ and $q= 2^{\alpha}3^{\beta}5^\gamma 7^{\delta}$
Where $a + \alpha = 8; b+\beta = 4; ...etc.....$
But as $\gcd(p,q) = 0$ we either have $a=0$ or $\alpha = 0$ and $b =0$ or $\beta = 0$ etc.
In other word we have $p = 2^{0,8}3^{0,4}5^{0,2}7^{0,1}$ while $q=2^{8,0}3^{4,0}5^{2,0}7^{1,0}$.
So there is two choices for $a:  0$ or $8$; all or nothing.  Same for the other variables.  And $\alpha, \beta, $ etc are completely dependent $a,b,c,d$.
So there are $2^4 = 16$ options.

In hindsight, it be better to answer

If $pq = 10!$ then $q = \frac {10!}p$ so we only have to calculate howmany choices of $p$ there are.
$p = 2^a3^b5^c7^d$ and $q = 2^{8-a}3^{4-b}5^{2-c}7^{1-d}$.  But as $\gcd(p,q)=1$ for each variable either $a =0$ or $8-a=0$ etc so there are $2$ options only.
So $p = 2^{0|8}3^{0|4}5^{0|2}7^{0|1}$ so there are four choices each with $2$ options.
So there $2^4=16$ choices.
.....
Oh.... I overlooked $\frac pq < 1$.
Oh well.  There are $16$ ordered pairs total.  As $\gcd(p,q) = 1$ (or because $10!$ is not a perfect square) we can't have $p < q$ and so there are $\frac {16}2=8$ unordered pairs or in other words pairs where $p < q$.
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Actually in hind-hindsight it's better to answer as Eyob Tsegaye did and realize it's enought to know $10! =2^m3^n5^s7^t$ and $p$ and $q$ have those powers as "all or nothing" and not have to figure the actual prime factorization.
