Find the probability in the following case. Nine balls, numbered $1, 2, \ldots, 9$, are put randomly at 9 equally
spaced points on a circle, each point with a ball. Let $S$ be the sum
of the absolute values of the differences of the numbers of all two
neighboring balls. Find the probability of $S$ to be the minimum
value.

Remark: If one arrangement of the balls is congruent to
another after a rotation or a reflection, the two arrangements are
regarded as the same

I've tried making all the cases. Like fixing a ball and arrangement of others. But it consumed a lot of time as there will be $ \frac{^8P_8}{2}$ cases. How could I solve this?
 A: $9$ balls with different numbers are placed at $9$ equally
spaced points on a circle, one point for one ball. This is equivalent to
a circular arrangement of $9$ distinct elements on a circle. Thus there
are $8!$ arrangements. Considering the reflections, there are $\frac{8!}{2}$
essentially different arrangements.
Next, we calculate the number of arrangements, which make $S$
the minimum. Along the circle there are two routes from $1$ to $9$, the
major arc and the minor arc. For each of them, let $x_1 , x_2,\ldots , x_k$ be
the numbers of the successive balls on the arc, then
$$|1-x_1|+|x_1-x_2|+\dotsb |x_k-x_9|\geq |(1-x_1)+(x_1-x_2)+\dotsb (x_k-x_9)|=|1-9|=8.$$
The equality occurs if and only if $1 < x_1 < x_2 <\ldots < x_k < 9$, i. e. the numbers of the balls on each route is increasing from $1$ to $9$. Therefore, $S_\min = 2 \times 8 = 16.$
From the above analysis, when the numbers of the balls
$\{1 , x_1 , x_2 , \ldots , x_k , 9 \}$ on each arc are fixed, the arrangement which
gets the minimum value is uniquely determined. Divide the set of $7$
balls $\{2, 3,\ldots, 8\}$ into two subsets, then the subset which contains
less elements has
$$C_{0,7}+C_{1,7}+C_{2,7}+C_{3,7}=2^6\,\text{cases}$$
Each case corresponds to a unique arrangement, which achieves the minimum value of $S$. Thus, the number of the arrangements when $S$ takes the minimun is
$2^6$  and the corresponding probability is
$$P=\frac{2^6}{\frac{8!}2}=\frac1{315}.$$
A: The minimum adjacent difference is 16. Its at least $16$, if we consider the path from $1 \to 9$ in both clockwise and anticlockwise directions. The sum of adjacent differences need to be at least $8$ in both ranges. We can construct $16$ simply with 123456789. Notice that effectively, this says the numbers in both directions towards 9 is monotone. Therefore you only need to think about the set of elements on the clockwise and anticlockwise arc. There are $2^7$ possible splits. We divide by $2$ after that to account for reflection.
