How many ways are there to arrange $8$ girls and $25$ boys to sit around a table so that there are at least $2$ boys between any pair of girls? If a sitting plan can be simply rotated to match another one, these two are treated as the same.
There are $8$ intervals between these girls. Let the number of boys in each of these intervals be $x_i$, $i=1,\ 2,\ \cdots,\ 8$, then $$x_1 + x_2+ \cdots + x_8=25$$
conditioned on $x_i\ge 2$. Let $y_i=x_i-2\ge 0$, then $$y_1+y_2 + \cdots + y_8=9$$
There are $\binom{16}{7}$ non-negative solutions to the above equation.
Meanwhile, there are $(8-1)!=7!$ ways to arrange $8$ girls in a round manner and $25!$ ways to arrange these boys. Hence, the final answer is $$\binom{16}{7}(7!)(25!)=\boxed{\frac{16!\cdot 25!}{9!}}$$