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)
 A: Since
\begin{align*}
720 & = 2 \cdot 360\\
    & = 2 \cdot 2 \cdot 180\\
    & = 2 \cdot 2 \cdot 2 \cdot 90\\
    & = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 45\\
    & = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 15\\
    & = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5\\
    & = 2^4 \cdot 3^2 \cdot 5
\end{align*}
each factor must be of the form $2^a3^b5^c$.
Suppose the factors are $2^{a_1}3^{b_1}5^{c_1}$ and $2^{a_2}3^{b_2}5^{c_2}$.  Observe that
\begin{align*}
a_1 + a_2 & = 4 \tag{1}\\
b_1 + b_2 & = 2 \tag{2}\\
c_1 + c_2 & = 1 \tag{3}
\end{align*}
are equations in the nonnegative integers.
The equation
$$x_1 + x_2 = n$$
has $n + 1$ solutions in the nonnegative integers, namely $$(n, 0), (n - 1, 1), (n - 2, 2), \ldots, (2, n - 2), (1, n - 1), (0, n)$$
Hence, equation 1 has five solutions, equation 2 has three solutions, and equation 2 has two solutions.  Hence, there are $5 \cdot 3 \cdot 2 = 30$ ordered pairs of factors with product $720$.  Since $720$ is not a perfect square, each ordered pair consists of two distinct numbers.  Since we do not care about the order of the factors, we must divide this result by two.  Thus, the number of ways to decompose $720$ into two positive integer factors is
$$\frac{5 \cdot 3 \cdot 2}{2} = 15$$
A: 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 $\frac{30}{2}=15$
