
4. Suppose 40 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 40 objects in $\binom{40}{3}$ ways. Among theese choices all would be together in 40 cases; exactly two will be together in $40 \times 36$ cases. Thus three objects can be chosen such that no two adjacent in $\binom{40}{3}-40-(40 \times 36)$ ways. Among these, furthrer, two objects will be diametrically opposite in 20 ways and the third would be on either semicircle in a non adjacent portion in $40-6=34$ ways. Thus required number is

$$
\binom{40}{3}-40-(40 \times 36)-(20 \times 34)=7720
$$
