I have found the answer to this RMO problem that baffled everyone

There are ${}^{32}C_3= 4960$ triples altogether.

Of these, $16 \times 30 = 480$ contain diametrically opposite points ( $16$ choices for the diametrically opposite points, and $30$ choices for the other point).

A total of $32\times28 + 32 = 32 \times 29 = 928$ triples contain adjacent points ( $32 \times 28$ of these triples have just two consecutive points, while the other $32$ have three consecutive points).

A total of $16\times4 =64$ triples contain adjacent points and diametrically opposite points.

Thus $4960 - 480 - 928 + 64 = 3616$ triples contain neither adjacent points nor diametrically opposite points.

Arrange all things in a straight line $a_{1},a_{2}...….a_{32}$

See the following arrangement

$P \boxed{a} Q \boxed{b} R \boxed{c} S$

Here a, b, c are any three things P, Q, R, S are no of things left in their middle and aside.

Here $P+Q+R+S =29$ where $P, S \geq 0 \& Q, R \geq 1$

So the no. of solutions of this equation are $^{27+4-1}C_{3}=4060$ . Now here we had taken them in a line.

When arranged in a circle $a_{1}$ and $a_{32}$ can't be together so we will remove the cases in which both of them are together, there are 28 cases to be removed.

So now we get 4060-28 = 4032 cases

Now we have to remove the cases of diametrically opposite things.

Here two things $a_{p}; a_{q}$ p>q are diametrically opposite if $p-q=16$ we will get 16 pairs with 26 arrangements corresponding to every pair so total cases are 26×16=416

Finally the answer comes out to be 4032-416=3616

Total cases are $32c3 = 4960$

Cases to be subtracted:

1) All 3 adjacent = 32

2) 2 adjacent and third one not adjacent to both= 32*28 = 896

3) 2 diametrically opposite and third one not adjacent to both = 16*26 = 416

So, answer is $4960 - 32 - 896 - 416 = 3616.$