
4. Suppose 32 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 32 objects in $\binom{32}{3}$ ways. Among these choices all would be together in 32 cases; exactly two will be together in $32 \times 28$ cases. Thus three objects can be chosen such that no two adjacent in $\binom{32}{3}-32-(32 \times 28)$ ways. Among these, furthrer, two objects will be diametrically opposite in 16 ways and the third would be on either semicircle in a non adjacent portion in $32-6=26$ ways. Thus required number is

$$
\binom{32}{3}-32-(32 \times 28)-(16 \times 26)=3616
$$
