Suppose $$10$$ objects are placed along a circle at equal distances. The number of ways can three objects be chosen from among them so that no two of the chosen objects are adjacent or diametrically opposite is $$\underline{\qquad}$$.

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 $$\frac{10\mathbin{\times}9}{3}=30$$.