Ten objects are placed at equal distances around a circle. The number of ways to choose three objects so that no two chosen objects are adjacent or diametrically opposite is

Fix one chosen object. Of the remaining positions, its two neighbours and its opposite position are forbidden, and counting the allowable pairs among the other six positions gives $$9$$ choices. Rotating the fixed choice gives $$10\mathbin{\times}9$$ counts, but every valid triple is counted once for each of its three objects. Thus the number of triples is $$\frac{10\mathbin{\times}9}{3}=30$$.