One can chose 3 objects out of 32 objects in (^{32}_{3}) ways. Among these choices all would be together in 32 cases; exactly two will be together in 32 x 28 cases. Thus three object can be chosen such that no two adjacent in (^{32}_{3}) - 32 - (32 x 28) ways. Among these, further, two objects will be diametrically opposite in 16 ways and third would be on either semicircle in a non adjacent portion in 32 – 6 = 26 ways. Thus required number is

(^{32}_{3}) - 32 - (32 x 28) - (16 x 26) = 3616