Example: 78 girls and 25 boys stand in a circle, with at least 2 boys standing between every two girls. There are $\qquad$ different ways of arranging them. (Arrangements that are the same when the circle is rotated are considered identical)
Solution One: Assume one of the girls is $A$. For any circular arrangement that meets the requirements, let's arrange them in a straight line starting from $A$ in a clockwise direction as $A B C D \cdots$ (as shown in Figure $1-3$). Now, let $\bigcirc$ represent the position of a girl and $\times$ represent the position of a boy. In the arrangement $A B C D \cdots$, each $\bigcirc$ is followed by at least two $\times$. Let each $\bigcirc$ "absorb" the two $\times$ immediately to its right, and represent this as $\oplus$. Thus, each $\bigcirc, \times$ arrangement corresponds to an $\oplus, \times$ arrangement.
$\bigcirc \times \times \times \bigcirc \times \times \times \times \bigcirc \times \times \times \times \times \bigcirc \times \times \cdots$
$\oplus \times$
$\oplus \times \times$
$\oplus \times \times \times$
$\oplus \cdots$

The latter arrangement consists of 8 " $\oplus$ " and $25-16=9$ " $\times$ ", totaling 17 elements, with $\oplus$ as the first element. The number of such arrangements is $C_{7+9}^{7}=C_{16}^{7}$. Therefore, the number of arrangements of boys and girls is $C_{16}^{7}$. With $A$ fixed in the first position, the girls can be arranged in $7!$ ways, and the boys can be arranged in $25!$ ways. Hence, the total number of arrangements is $C_{16}^{7} \cdot 7! \cdot 25! = 16! \cdot 25! / 9!$.