$$
3^{r} \cdot 7^{s}=[a, b, c]=[a, b, d]=[a, c, d]=[b, c, d] \text {. }
$$

Find the number of such quadruples (answer as a function of $r, s$).
The answer is $\left(1+4 r+6 r^{2}\right)\left(1+4 s+6 s^{2}\right)$.
In fact, each of $a, b, c, d$ is in the form of $3^{m} \cdot 7^{n}$, where $m \in\{0,1,2, \cdots, r\}, n \in$ $\{0,1,2, \cdots, s\}$. Moreover, at least two of $a, b, c, d$ have $m=r$, and at least two have $n=s$. When all of $a, b, c, d$ have $m=r$, there is only one way to choose. When one of $a, b, c, d$ has $m \in\{0, 1, \cdots, r-1\}$, and the other three have $m=r$, there are $4 r$ ways to choose. When two of $a, b, c, d$ have $m \in\{0,1, \cdots, r-1\}$, and the other two have $m=r$, there are $\mathrm{C}_{4}^{2} \cdot r^{2}=6 r^{2}$ ways to choose. Thus, there are $1+4 r+6 r^{2}$ ways to choose $m$, and similarly, there are $1+4 s+6 s^{2}$ ways to choose $n$. The total number of ways is $(1+$ $\left.4 r+6 r^{2}\right)\left(1+4 s+6 s^{2}\right)$.