$a,b,c$ are three distinct natural numbers. Then how many ordered triplets $(a,b,c)$ will exist such that L.C.M (a,b,c) = 144. Let $a,b,c$ be three distinct natural numbers. Then how many ordered triplets $(a,b,c)$ will exist such that L.C.M (a,b,c) = 144.
Here's how I proceeded, 
144=$(2^4)(3^2)$, so 144 has 15 factors(1 and 144 included). Let $F$ be the set of these factors. Then $a,b,c\in{F}$. I can't just select any 3 elements of $F$ as their LCM need not necessarily be 144. Is there any sort of further condition which may be applied on $a,b,c$ that will help me solve this problem.

At least one of them must have $2^4$ and at least one of them must have $3^2$. 
Assign $3^2$ to any one and what choices do you have with others? Proceed similarly with $2^4$
