Problem: If 240 equals $a b c$, the product of positive integers $a, b$ and $c$, compute the number of distinct ordered triples $(a, b, c)$ such that $a$ is a multiple of $2, b$ is a multiple of 3 and $c$ is a multiple of 5. Solution: The answer is $10$. We can dispose of the annoying division condition by allocating divisors before we count. That is, we know we must have $a = 2a_{0}, b=3b_{0}, c=5c_{0}$ where $a_{0},b_{0},c_{0} \in \mathbb{N}$. Hence, $abc = 30a_{0}b_{0}c_{0} = 240 \iff a_{0}b_{0}c_{0} = 2^3$. Now, we ask, how many ways are there to distribute three factors of $2$ among $a_{0},b_{0},c_{0}$. This is an application of distributions and is just $\binom{3 + 3-1}{3 - 1} = 10$