**Problem: ** Suppose 36 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite.

**Discussion:**

Without any restriction 3 points can be chosen in \(\binom{36}{3} \) ways.

We will delete the cases which are not allowed.

** Case 1 ** – all three points are adjacent.

Exactly 36 cases are possible, with one such triangle for each vertex.

** Case 2 ** – exactly two points are adjacent

First we choose one of the 36 such adjacent pairs in \(\binom{36}{1} \) 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 \(\binom{32}{1} \) ways (36 – two points we have already selected – two points adjacent to these two).

Hence total count is \(\binom{36}{1} \times \binom{32}{1} \)

** Case 3 ** – No two points are adjacent but two of them are diametrically opposite.

We choose a diameter in 18 ways (since there are 36 points equally spaced, there will be 18 diameters).

The third point chosen cannot be adjacent to any one of the end points of this chosen diameter. Hence we have \(\binom{30}{1} \) ways.

Hence total count is \(18 \times \binom{30}{1} \)

Therefore the total number of favorable cases are: \(\binom{36}{3} – 36 – \binom{36}{1} \times \binom{32}{1} – 18 \times \binom{30}{1} = 5412 \)

## Chatuspathi:

**What is this topic:**Combinatorics**What are some of the associated concept:**Compliment rule of counting**Where can learn these topics:**Cheenta**Book Suggestions:**Principles of Combinatorics by Chuang Chong Chen

shouldnt it be 18C30 instead of 18C32?

After taking any two diametrically opp points, we have 34 points left

but for each diametrically opp point chosen, we have 2 adjacent points that we cant choose

therefore 34-4=30

That is true. Corrected. Thank you.

I meant 18*30C1.

how am i wrong? at first i take 33 objects round a table.there are 33 gaps between them.so 3 gaps can be chosen in 33C3 ways.in each of such cases no two of 3 are adjuscent.so no of ways is 33C3=5416.then I substruct 540 cases(as discussed here).so required no of ways is 5416-540=4916.what is wrong with this argument????

one of the issues that I right away notice that you have not considered the restriction regarding diametrically opposite points.

Sir, in MP Region RMO, in this question it is 28 objects in place of 36 objects. I could not solve the problem. Was it misprint or this question has a solution with given 28 objects?

Rudr, it is always good to heat from you. Hopefully you had good test.

Of course it has solution for 28 objects

Yes Sir, Hoping to get through. Thanks

Please solve this problem :

PQRS is a square. T is any point on PR. Perpendiculars TA and TB are drawn on PS and PQ respectively.TW is drawn perpendicular to AQ and RZ is drawn perpendicular to AQ. Prove that AW=ZQ.