
4. Suppose 36 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 36 objects in $\binom{36}{3}$ ways. Among these choices all would be together in 36 cases; exactly two will be together in $36 \times 32$ cases. Thus three objects can be chosen such that no two adjacent in $\binom{36}{3}-36-(36 \times 32)$ ways. Among these, furthrer, two objects will be diametrically opposite in 18 ways and the third would be on either semicircle in a non adjacent portion in $36-6=30$ ways. Thus required number is

$$
\binom{36}{3}-36-(36 \times 32)-(18 \times 30)=5412
$$
