Example 19 (1990 National High School League Question) 8 girls and 25 boys stand in a circle, with at least two boys standing between any two girls. How many different arrangements are there (arrangements that coincide by rotating the circle are considered the same)?
Fix one of the girls as $A$, for any circular arrangement that meets the requirement, transform it into a linear arrangement starting from $A$ in a clockwise direction. Since each girl is followed by at least two boys, we can map each girl along with the two boys immediately following her to a black ball, and the remaining 9 boys to white balls. Thus, without considering the differences between girls and between boys, this mapping is a bijection. The correspondence between the arrangements is also a bijection. Therefore, the problem is converted to calculating the number of all arrangements of 8 black balls and 9 white balls where the first ball is black. Clearly, there are $C_{16}$ different arrangements.

For each arrangement of black and white balls, each black ball corresponds to a group of three, where the first is a girl and the next two are boys. Arrange the 7 girls other than $A$ in the 7 groups of three, which can be done in $7!$ ways; arrange the 25 boys into 25 positions, which can be done in 25! ways. By the multiplication principle, the total number of arrangements is $\frac{16! \cdot 25!}{9!}$.