<s>[INST] How to find the numbers of sets of positive numbers $\{a,b,c\}$ such that $(a)(b)(c) =2^{4}3^{5}5^{2}7^{3}$? How to find the numbers of sets of positive numbers $\{a,b,c\}$ such that $(a)(b)(c) =2^{4}3^{5}5^{2}7^{3}$ ? 
$$\binom{4+2}{2} \binom{5+2}{2}  \binom{2+2}{2} \binom{3+2}{2}$$
This would give the number of tuples $(a,b,c)$. But isolating the individual sets $\{a,b,c\}$ is harder as there would be duplicates. Any ideas how to do it?
 [/INST] Let $S = \{(a,b,c) \colon abc = 2^4 3^5 5^2 7^3\}$. Let's split it into few smaller sets. $S_1 = \{(a,b,c) \in S \colon a \neq b;\, b \neq c;\, a \neq c\}$, $S_2 = \{(a,b,c) \in S \colon a=b;\, b \neq c\}$ and let $S_3, S_4$ be similar to $S_2$ only with equal pairs $a=c$ and $b=c$. It's easy to notice that $S = S_1 \cup S_2 \cup S_3 \cup S_4$, because the case $a=b=c$ is impossible for our condition (otherwise all powers should be divisible by 3). Also $|S_2| = |S_3| = |S_4|$. 
The answer for the problem is ${|S_1| \over 6}$, because each possible set of three elements contributes to $S_1$ six times. We know $|S|$ hence all we have to do is find $|S_2|$. It's quite easy, let's consider some element of $S_2$. $(a,a,b) = (2^{a_1} 3^{a_2} 5^{a_3} 7^{a_4}, a, 2^{b_1} 3^{b_2} 5^{b_3} 7^{b_4})$ and from our condition 
$$\left\{ \begin{array}{l} 2a_1 + b_1 = 4\\
                           2a_2 + b_2 = 5 \\
                           2a_3 + b_3 = 2\\
                           2a_4 + b_4 = 3
\end{array} \right.$$
Each condition is independent from the others and there are $3$ possibilities to satisfy the first condition, $3$ for the second, $2$ for the third and $2$ for the fourth. So $|S_2| = 3 \cdot 3 \cdot 2 \cdot 2$. $|S_1| = |S| - 3 |S_2|$. We're done.
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