6. 8 girls and 25 boys stand in a circle, with at least two boys standing between any two girls. Therefore, there are $\qquad$ different arrangements (arrangements that coincide by just rotating the circle are considered the same).
6. Suppose one of the girls is $A$, for any circular arrangement that meets the requirements, starting from $A$ and moving clockwise, it forms a straight line, as shown in the diagram below:
$\longleftrightarrow A B C D \cdots$

Now, let $\bigcirc$ represent the position of a girl and $\times$ represent the position of a boy, then after each $\bigcirc$ there are at least two $\times$. Let each $\bigcirc$ "absorb" the two $\times$ immediately following it, and draw it as a థ, then each $\bigcirc$, $\times$ arrangement corresponds to a $\oplus$, $\times$ arrangement: $\square$
$$
\underbrace{\bigcirc \times \times}_{\oplus} \quad \underbrace{\bigcirc \times \times}_{\oplus} \times \underset{\oplus}{\bigcirc} \underbrace{\bigcirc \times \times}_{\oplus} \quad \cdots
$$

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., $C_{16}^{7}$.

The above shows that the number of ways to arrange the positions of boys and girls is $C_{16}^{7}$. 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
$$
C_{16}^{7} 7!25!=\frac{16!25!}{9!} .
$$