An Interesting riddle on Rational's and Primes We are supposed to find all rational numbers $\frac{m}{n}$  such that the following set of conditions are true:
$$1)0<\frac{m}{n}<1$$
$$2)H.C.F [m, n] =1$$
$$3)m⋅n = 30!$$
Here is my attempt at it:

I used Legendre's formula to find out powers of prime factors involved so I arrived at the following: $2^{26}, 3^{14},5^{7},7^{4},11^{2},13^1,17^1,19^1,23^1,29^1$ now to make m, n and to make them coprime I decided to make groups of these factors(prime numbers) without distributing the powers of these primes i.e. I am taking whole of $2^{26}$ in a group not taking $2^{13}$ in one and $2^{13}$ in other,as that will result in H.C.F being 2 not 1. hence we need to distribute these 10 factors in groups of 2. We can make these many groups by the following scheme : $\binom{10}0+\binom{10}9+\binom{10}8+\binom{10}7+\binom{10}6+\binom{10}5$ this is because we can choose 10 factors at once and call it n (here m being 1) also we can choose 9 factors and in each choice 1 of the 2 pairs will result in greater number so accordingly for each $\binom{10}9$ 1 possibility would be chosen so we can keep doing this till we reach $\binom{10}5$ now furthur choosing will not result in new cases being formed just same groups will be made again. Hence this should be the answer.

 A: (1) In your list of prime powers in the first paragraph you left out $7^4$ (though you seemed to have counted it in your total of $10$ primes).
(2) The ${{10}\choose{5}}$ term should be divided by $2$, otherwise you count the answers with five prime factors in each part twice. (For instance once you select $2, 3, 5, 7, 11$ as the primes, and once you select $13, 17, 19, 23, 29$ as the primes, but that gives the same values for $m$ and $n$. (With this correction, you should get an answer consistent with the comment of Summer Child.)
A: I don't get the method by which you calculate the number of solutions.  There are exactly $2^{10-1} = 512$ solutions:  $30!$ factors into 10 prime powers:
$$30!=\prod_{k=1}^{10}p_k^{m_k}$$
where $p_k$ are primes and $m_k\in\Bbb N^+$. In order to enumerate solutions, consider all bit strings $S\in\{0,1\}^9$ of length 9 and set
$$ a = a(S) = p_1^{m_1} \prod_{k=2}^{10}p_k^{m_k\cdot b_k}$$
where $b_k\in\{0,1\}$ is the $k-1$-th bit of $S$. Then define
$$\begin{align}
m &= \min(a, 30!/a) \\
n &= \max(a, 30!/a) \\
\end{align}$$
For example, when we order the primes such that $p_k < p_{k+1}$:
$$\begin{align}
a((0,0,0,0,0,0,0,0,0)) &= 2^{26} \\
a((1,0,0,0,0,0,0,0,0)) &= 2^{26} \cdot 3^{14} \\
a((0,1,0,0,0,0,0,0,0)) &= 2^{26} \cdot 5^{7} \\
a((1,1,0,0,0,0,0,0,0)) &= 2^{26} \cdot 3^{14} \cdot 5^{7} \\
&\ \ \vdots \\
a((1,1,1,1,1,1,1,1,1)) &= 2^{26} \cdot \underbrace{3^{14} \cdot 5^{7} \cdot 7^4\cdots 29}_{\textstyle 9\text{ factors}} =30!\\
\end{align}$$
Then we have that $\gcd(m,n) = 1$ and $0<m<n$ and $m\cdot n=30!$ by construction. $m\neq n$ because $30!$ is not a perfect square, and $m(S_1)\neq m(S_2)$ whenever $S_1\neq S_2$ because prime factorization is unique (the order of the primes is fixed).

You can also calculate the numer of solutions as follows:  $30!$ factors into 10 prime powers. You have $2^{10}$ ways to select whether a prime power shall be a factor of $m$ or not, and the other way rounf for $n$, i.e. $n=30!/m$.  Exactly half of all the $m$'s will satisfy $m>n$ and hence are no valid solutions.  The other half of the $m$'s satisfy $m<n$ and are valid.  Thus, there are $2^{10}/2$ valid pairs $(m,n)$.
