Question 192: Nine small balls numbered $1, 2, \ldots, 9$ are randomly placed on nine equally spaced points on a circle, with one ball on each point. Let $S$ be the sum of the absolute differences of the numbers on all adjacent balls on the circle. Find the probability of the arrangement that minimizes $S$ (Note: If one arrangement can be made to coincide with another by rotation or reflection, they are considered the same arrangement).
Question 192, Answer: Place nine differently numbered small balls on nine equally divided points on a circle, with one ball per point. This is equivalent to a circular permutation of nine different elements, with a total of 8! ways to place them. Considering the factor of flipping, there are $\frac{8!}{2}$ essentially different ways. Next, we need to find the number of ways to achieve the minimum value of $\mathrm{S}$.

On the circle, from 1 to 9, there are two paths: the major arc and the minor arc. For any one of these paths, let $x_{1}, x_{2}, \ldots, x_{k}$ be the numbers of the balls sequentially arranged on this arc. Then $\left|1-x_{1}\right|+\left|x_{1}-x_{2}\right|+\ldots+\left|x_{k}-9\right| \geq \mid\left(1-x_{1}\right)+\left(x_{1}-x_{2}\right)+\ldots+\left(x_{k}-9\right) \mid=8$, equality holds if and only if $1<x_{1}<x_{2}<\ldots<x_{k}<9$, i.e., the numbers of the balls on each arc are arranged in increasing order from 1 to 9. Therefore, the minimum value of $S$ is $2 \times 8=16$.

On the other hand, by arbitrarily partitioning $\{2,3, \ldots, 8\}$ into two subsets $A$ and $B$, and arranging the elements of each subset in increasing order, we exactly correspond to the two paths that make $S=16$. Hence, the number of ways to make $S=16$ is $\frac{2^{7}}{2}=2^{6}$. Therefore, the probability of the ways that make $S$ reach its minimum value is $\frac{2^{6}}{\frac{8!}{2}}=\frac{1}{315}$.