Question from a competitive exam:
Find the number of possible ordered triples $(a,b,c)$ such that $abc=108$.
Attempt at solution:
Factorizing $108$ gives $3^3$$2^2$. So an ordered triple will be of the form $(3^{a_1} 2^{b_1} ,3^{a_2} 2^{b_2} ,3^{a_3}2^{b_3})$ where $a_1+a_2+a_3=3$ and $b_1+b_2+b_3=2$. I am unsure how to proceed from here.
As a side note, I gave the answer as $18$ based on flawed logic.
From here you can count directly the number of possibilities for $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$. For the $b_i$ the only options are to have $2,0,0$ in some order ($3$ ways) or $1,1,0$ in some order ($3$ ways), so there are $6$ possibilities for the $b_i$. You can then work out the number of possibilities for the $a_i$ using similar methods, and then multiply by the $6$ possible $b_i$ to get the answer.
For larger powers where it is harder to do this by hand, there is a general formula which comes from a "stars and bars" argument.