
4. 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?


Solution: One can choose 3 objects out of 28 objects in $\binom{28}{3}$ 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 $\binom{28}{3}-28-(28 \times 24)$ 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

$$
\binom{28}{3}-28-(28 \times 24)-(14 \times 22)=2268
$$
