We can select 3 objects from 36 objects in $36 C3$ ways. next we try to eliminate the cases that are not given. CASE 1- all three points are adjacent- we see that there are exactly $ 36$ possible cases CASE 2- any and exactly two points are adjacent- 1 pair out of 36 adjacent pairs in $ 36C1$ and one point out of the 32 remaining points(2 points of the pair and their adjacent 2 points) in $32C1$ Thus in this case $(36C1 * 32C1)$ CASE 3- no two are adjacent but two are diametrically opposite- 18 ways to choose the diameter and the third point cannot be adjacent to any of the chosen diameter So $30 C1$ (since two points on the diametric ends and each of their two adjacent points,36-2-2-2=30) Hence we calculate the total now, $ 36C3 -36-36C2*32C1- 18*30C1 $ =5412 Ans.