# West Bengal RMO 2015 Problem 4 Solution - 36 objects in a Circle

The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 4 Solution has been written for RMO preparation series.

## 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 I.S.I. & C.M.I. course, Cheenta Math Olympiad Program, discuss these topics in the ‘Number Theory’ module.
• Book Suggestions: Principles of Combinatorics by Chuang Chong Chen

### 11 comments on “West Bengal RMO 2015 Problem 4 Solution - 36 objects in a Circle”

1. […] Suppose objects are placed along a circle at equal distances. In how many ways can objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite. SOLUTION: Here […]

2. . says:

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

1. . says:

I meant 18*30C1.

3. ANANTA MUKHERJEE says:

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????

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

1. Rudr Tiwari says:

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?

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

Of course it has solution for 28 objects

4. Rudr Tiwari says:

Yes Sir, Hoping to get through. Thanks

5. Ankur says: