Example: 78 girls and 25 boys stand in a circle, with at least two boys standing between any two girls. How many different arrangements are there?
$(1990$, National High School Mathematics Competition)
Solution: Arrangements that can coincide after rotation should be considered as the same arrangement. Therefore, we can consider fixing one girl $A$ and stretching the circle into a straight line in a clockwise direction. Let $a$ and $b$ represent girls and boys, respectively. First, choose 16 boys from 25 to form " $A b b$ " and 7 groups of " $a b b$ ", which can be done in $\mathrm{C}_{25}^{16} \mathrm{P}_{16}^{16}$ ways. Then, arrange the remaining 9 boys and the 7 groups of " $a b b$ " (excluding " $A b b$ ") as 16 groups in any order, which can be done in $\mathrm{P}_{16}^{16}$ ways. Therefore, the number of different arrangements is
$$
\mathrm{C}_{25}^{16} \cdot \mathrm{P}_{16}^{16} \cdot \mathrm{P}_{16}^{16}=\frac{25!16!}{9!} \text {. }
$$

Note. The method used in this example, which involves first constructing "small wholes" that meet the conditions, is widely applicable and worth noting.