Select Page

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