How many ways can $720$ be decomposed into a product of two positive integers? 
How many ways can $720$ be decomposed into a product of two positive integers?

My solution: There are $5 \cdot 3 \cdot 2 = 30$ ways to choose the exponents a, b, c, such that $2^a \cdot 3^b \cdot 5^c= 720$. Soon there are $30$ dividers. As there are $30$ dividers, there are $\frac{30}{2} = 15$ different products in the form $xy=720$
Question: We want $ab$ such that $a,b \in \{2^5 \cdot 3^3 \cdot 5^2\}$, that is, we want combinations of these possible values of $a$ and $b$, such that $ab=720$. How could I solve this problem following this line of reasoning? A different solution from mine (Using counting principles)

I believe this may be another approach:
We want $ab=720$. Since we have $2^5 \cdot 3^3 \cdot 5^2$ and we want this to form the product of 2 primes such that it results in $720$, then:
$$\binom{5}{2} \cdot \binom{3}{2} \cdot  \binom{2}{2}= 30.$$ But since we ended up counting the possibilities twice (for example: if $a=2$, then $b=360$; if $b=2$, then $a=360$), then we have $\frac{30}{2}=15$
