suppose 28 objects are placed along a circle at equal distancces.Number of ways in which 3 objects can be chosen from them so that no two of three objects are adjacent nor diametrically opposite is...........

Without any restriction 3 points can be chosen in 28C3 ways.
We will delete the cases which are not allowed.

Case 1 – all three points are adjacent.
Exactly 28 cases are possible, with one such triangle for each vertex.

Case 2 – exactly two points are adjacent
First we choose one of the 28 such adjacent pairs in 28C1 ways. Next the third point is chosen such that it is not adjacent to any one of the two chosen points. Hence it can be done in 24C1 ways (28 – two points we have already selected – two points adjacent to these two).
Hence total count is 28C1 * 24C1

Case 3 – No two points are adjacent but two of them are diametrically opposite.
We choose a diameter in 14 ways (since there are 28 points equally spaced, there will be 14 diameters).
The third point chosen cannot be adjacent to any one of the end points of this chosen diameter. Hence we have 22C1 ways.
Hence total count is 14 * 22C1
Therefore the total number of favorable cases are:
28C3 - 28 - 24C1 * 28C1 - 14 * 22C1