6. 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)?
$$
\frac{16!25!}{9!}
$$
6. 【Analysis and Solution】Assume one of the girls is $A$, for any circular arrangement that meets the requirements, start from $A$ and walk clockwise to form a straight line, as shown in the diagram below:
$\leftrightarrow A B C D \cdots$
Now, let $O$ represent the position where a girl stands, and $\times$ represent the position where a boy stands, then after each $O$ there are at least two $\times$. If each $O$ "absorbs" the two $\times$ immediately after it, and is drawn as a $\oplus$, then each $O, x$ arrangement corresponds to a $\oplus, x$ arrangement:
$\underbrace{\mathrm{O} \times x}_{\oplus}$
$\times$
$\underbrace{\mathrm{O} x}_{\oplus}$ $\qquad$ $\cdots$
The number of the latter type of arrangement is clearly the number of ways to choose 7 positions from $(8+25)-2 \times 8-1=16$ positions, i.e., $\mathrm{C}_{16}^{7}$.
The above shows that the number of ways to arrange the positions of boys and girls is $\mathrm{C}_{16}^{7}$. For each position arrangement, the girls can stand in $7!$ ways (with $A$ fixed at the first position), and the boys can stand in 25! ways, so the total number of arrangements is
$$
C_{16}^{7} 7!25!=\frac{16!25!}{9!}
$$