How many pairs of positive integers $x, y, x \leqslant y$, satisfy $(x, y)=5!$ and $[x, y]=50!$?
1. Let $p_{1}, p_{2}, \cdots, p_{12}$ represent the 12 prime numbers from 7 to 47 in increasing order.
Then, $5!=2^{3} \cdot 3^{2} \cdot 5^{1} \cdot p^{3} \cdot f_{2}^{\prime \prime} \cdot \cdots \cdot f_{12}^{\prime}$.
For 50! $=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{*_{3}} \cdot \hat{o}_{1}^{b_{1}}=p_{2}^{b_{2}} \cdot \cdots \cdot p_{12}^{b_{12}}$.
It is given that $2^{4}, 3^{2}, 5^{2}, \dot{p}_{1}, p_{2}, \cdots, p_{12}$ all divide 501, hence the prime numbers $2,3,5, p_{1}, p_{2}, \cdots, p_{12}$ have different exponents in 50! compared to their exponents in 5!.
$$
\begin{array}{l}
\text { Also, since } x|50!, y| 50!, \text { thus } x, y \text { have the form } \\
x=2^{n_{1}} \cdot 3^{m_{2}} \cdots \cdot p_{12}^{n_{12}}, \\
y=2^{m_{1}} \cdot 3^{m_{2}} \cdots \cdots p_{12}^{m_{15}} .
\end{array}
$$

Then, the exponent of the $i$-th prime in the factorization of 50! is $\max \left(n_{i}, m_{i}\right)$, and the exponent of the $i$-th prime in the factorization of 5! is $\min \left(n_{i}, m_{i}\right)$.

From the above, the exponents of the prime numbers $2,3,5, p_{1}, p_{2}, \cdots, p_{12}$ are different in 5! and 50!. Therefore, there are $2^{15}$ choices for $x$, and only half of these choices make $x$ not exceed $y$ (since for each choice of $x$, $y$ is determined, and either $x<y$ or $y<x$). Hence, the number of such integers is $\frac{2^{15}}{2}=2^{14}$.