Suppose 28 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite?

One can choose 3 objects out of 28 objects in  $\left(\begin{array}{c}28\\ 3\end{array}\right)$ ways. Among these choices all would be together in 28 cases; exactly two will be together in  $28\times 24$ cases. Thus three objects can be chosen such that no two adjacent in  $\left(\begin{array}{c}28\\ 3\end{array}\right)-28-(28\times 24)$ ways. Among these, furthrer, two objects will be diametrically opposite in 14 ways and the third would be on either semicircle in a non adjacent portion in  $28-6=22$ ways. Thus, required number is  $\left(\begin{array}{c}28\\ 3\end{array}\right)-28-(28\times 24)-(14\times 22)=2268$