Example 2-57 Find the number of permutations of the 8 letters A, B, C, D, E, F, G, H where only 4 elements are not in their original positions.
8 letters have only 4 not in their original positions, while the other 4 letters remain unchanged, which is equivalent to the derangement of 4 elements, the number of which is
$$
\begin{aligned}
4!\left(1-\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}\right) & =24\left(1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\right) \\
& =12-4+1=9
\end{aligned}
$$

Therefore, the number of permutations of 8 letters where 4 are not in their original positions is
$$
\begin{aligned}
C(8,4) \cdot 9 & =\frac{8!}{4!4!} \cdot 9=\frac{8 \cdot 7}{4} \cdot 6 \cdot 5 \\
& =7 \cdot 6 \cdot 5 \cdot 3=210 \cdot 3=630
\end{aligned}
$$