Problem: Find the number of ordered pairs of integers $(a, b)$ such that $a, b$ are divisors of 720 but $a b$ is not. Solution: Consider a complementary counting approach. Let $A$ be the set of **positive integral** ordered pairs $(a,b)$ such that $a\mid 720, b\mid 720$ but $ab \nmid 720$. It is clear that $ |A| = |\{(a,b) \in \mathbb{Z}^2: a,b \mid 720\}| - |\{(a,b) \in \mathbb{Z}^2: a,b,ab\mid 720\}|. $Since $720 = 2^{4}3^{2}5^1$ we know that the number of **positive** divisors of $720$ is $d(720) = (4+1)(2+1)(1+1) = 30$. Hence, the former set is cardinality $30^2 = 900$. On the other hand, to count the ladder set, we can count this by considering three boxes where the first box represents the factors that go in $a$, the second box denotes the factors that go in $b$, and the third box is the factors to discard. By the end of placing all the prime factors of $720$ among these boxes, we end up with $a,b$ such that both divide $720$ and - in particular - $ab$ also divides 720. Also, we can clearly see each arrangement of $(a,b)$ has a unique box representation (all of this is to say this method is a well-defined bijection and so counting this suffices). Hence, the latter set has cardinality $\binom{4 + 3-1}{3-1} \binom{2 + 3 - 1}{3 - 1} \binom{1 + 3-1}{3-1} = 15 \cdot 6 \cdot 3 = 270$. So, $|A| = 900-270 = 630$. Now, since the sign doesn't matter, each element of each ordered pair can be positive or negative, so the final answer is $4 \cdot 630 = 2520$.