An algebra problem by Anish Roy

One can choose 3 objects out of 32 objects in $32 \choose 3$ 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 ${32 \choose 3}-32-(32 \times 28)$ 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 ${32 \choose 3}-32-(32 \times 28)-(16 \times 26)=\boxed {3616}$