Example 27 (1997 Canadian Mathematical Olympiad) How many pairs of positive integers $x, y$ satisfy the condition $(x, y)=5!$ and $\lceil x, y\rceil=50!$?
Let the primes between 7 and 47, $7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$, be denoted as $p_{1}, p_{2}, \cdots, p_{12}$, then
$$
\begin{array}{l}
5!=2^{3} \cdot 3^{1} \cdot 5^{1} \cdot p_{1}^{0} \cdot p_{2}^{0} \cdot \cdots \cdot p_{12}^{0}, \\
50!=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot p_{1}^{b_{1}} \cdot p_{2}^{h_{2}} \cdot \cdots \cdot p_{12}^{b_{12}} .
\end{array}
$$

Since $x|50!$ and $y|50!$, $x$ and $y$ have the following forms:
$$
\begin{array}{l}
x=2^{n_{1}} \cdot 3^{n_{2}} \cdot 5^{n_{3}} \cdot p_{1}^{n_{4}} \cdots \cdots \cdot p_{12}^{n_{15}}, \\
y=2^{m_{1}} \cdot 3^{m_{2}} \cdot 5^{m_{3}} \cdot p_{1}^{m_{4}} \cdots \cdots \cdot p_{12}^{m_{5}} .
\end{array}
$$

where $\max \left\{n_{i}, m_{i}\right\}$ is the exponent of the $i$-th prime in $50!$, and $\min \left\{n_{i}, m_{i}\right\}$ is the exponent of the $i$-th prime in $5!$. Without considering the two possible cases of $xy$.
Thus, the number of positive integer pairs that satisfy the condition is $\frac{2^{15}}{2}=2^{14}$.