Solution:

We note that $54000 = 2^{4} \times 3^{3} \times 5^{3}$. Hence, we must have $a = 2^{a_{1}} 3^{a_{2}} 5^{a_{3}}$, $b = 2^{b_{1}} 3^{b_{2}} 5^{b_{3}}$, $c = 2^{c_{1}} 3^{c_{2}} 5^{c_{3}}$. We look at each prime factor individually:

- $4 a_{1} + 2 b_{1} + c_{1} = 4$ gives 4 solutions: $(1,0,0), (0,2,0), (0,1,2), (0,0,4)$
- $4 a_{2} + 2 b_{2} + c_{2} = 3$ and $4 a_{3} + 2 b_{3} + c_{3} = 3$ each give 2 solutions: $(0,1,1), (0,1,3)$.

Hence, we have a total of $4 \times 2 \times 2 = 16$ solutions.