combinatorics circular arrangement problem If $n$ distinct things are arranged in a circle, then what are the number of ways selecting three of these things so that no two of them are next to each other?

For $n\leq 5$, it is clear that it is impossible.
So let $n\geq 6$.
For the first object, there are $n$ possibilities.
Now 3 objects are forbidden (the chosen one and the distance 1 objects).
If as the second object you pick one of the 2 object in distance 2, for the last object there are $n-5$ possibilities. If as the second object you take one of the $n-5$ objects in distance $>2$, for the last objects there are $n - 6$ possibilities. So in total
$$
n\cdot (2(n-5) + (n-5)(n-6)) = n(n-5)(n-4)
$$
possibilities or, if you the order of the chosen object doesn't matter
$$
\frac{n(n-5)(n-4)}{6}
$$
possibilities.
