8. Determine the number of ordered quadruples $(a, b, c, d)$ of positive integers such that $a b c d=216$.
Answer: 400
Solution: Since $216=2^{3} 3^{3}$, any positive divisor of 216 must be of the form $2^{x} 3^{y}$ for some integers $x$ and $y$ with $0 \leq x, y \leq 3$. Thus, we set $a=2^{x_{1}} 3^{y_{1}}, b=2^{x_{2}} 3^{y_{2}}, c=2^{x_{3}} 3^{y_{3}}$ and $d=2^{x_{4}} 3^{y_{4}}$, where $0 \leq x_{i}, y_{i} \leq 3$ are integers for $i=1, \ldots, 4$. We compute
$$
2^{3} 3^{3}=216=a b c d=2^{x_{1}+x_{2}+x_{3}+x_{4}} 3^{y_{1}+y_{2}+y_{3}+y_{4}}
$$
so the number of such ordered quadruples $(a, b, c, d)$ is the number of ordered 8-tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}, y_{1}\right.$, $\left.y_{2}, y_{3}, y_{4}\right)$ of nonnegative integers such that $x_{1}+x_{2}+x_{3}+x_{4}=y_{1}+y_{2}+y_{3}+y_{4}=3$. By stars-and-bars, this number is $\binom{3+4-1}{4-1}^{2}=20^{2}=400$.