# Understand the problem

Consider the different ways to colour a cube with $6$ given colours such that each face will be given a single colour and all the six colours will be used. Define two such colourings to be equivalent if one will get from another just by rotation. Then what would be the number of inequivalent colourings?

##### Source of the problem

TIFR 2019 GS Part A, Problem 18

Combinatorics
Moderate
##### Suggested Book

Do you really need a hint? Try it first!

For example, first let us choose the colours on the two opposite sides in $^6C_2$ ways, then another two opposite sides in $^4C_2$ ways, and the leftover sides in $^2C_2$ ways. So, in total, we get $15 \times 6=90$, but think about over counting.
you’re choosing three pairs of oppositely placed colours, but you are overcounting because the same division into three pairs can be chosen in
$6$ different orders, being counted erroneously as distinct by you while. At the same time, you consider one colouring and its mirror image non-distinct (swapping two opposite colours is erroneously seen as the same choice). This gives a total overcounting by a factor of $\frac62=3$.
Another approach: There are $6!=720$ different ways of colouring without taking rotations into consideration. There are $24$ different rotations, and for any colouring, no two rotations give identical result. Therefore these $720$ colourings may be divided into groups of $24$ essentially equal colourings. There must be $30$ such groups.
Another approach: In a circular permutation, arrangements which may be obtained from one another by rotating the cube are considered identical. Place a colour on the bottom. It does not matter which. The top colour may be chosen in five ways. Place another colour facing you. Again, it does not matter which. The remaining three colours may be arranged relative to the colour facing you in $3!$ ways as we proceed clockwise around the cube. Hence, there are $5 \cdot 3! = 30$ admissible colourings.

# Connected Program at Cheenta

#### College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

# Similar Problems

## Sequences & Subsequences : IIT 2018 Problem 10

This problem appeared in IIT JAM 2018 whch pricisely reqiures concepts of sequences and subsequences from mathematical field real analysis

## Cyclic Groups & Subgroups : IIT 2018 Problem 1

This is an application abstract algebra question that appeared in IIT JAM 2018. The concept required is the cyclic groups , subgroups and proper subgroups.

## Acute angles between surfaces: IIT JAM 2018 Qn 6

This is an application analysis question that appeared in IIT JAM 2018. The concept required is the multivarible calculus and vector analysis.

## Finding Tangent plane: IIT JAM 2018 problem 5

What are we learning?Gradient is one of the key concepts of vector calculus. We will use this problem from IIT JAM 2018 will use these ideasUnderstand the problemThe tangent plane to the surface $latex z= \sqrt{x^2+3y^2}$ at (1,1,2) is given by $$x-3y+z=0$$...

## An excursion in Linear Algebra

Did you know Einstein badly needed linear algebra? We will begin from scratch in this open seminar and master useful tools on the way. The open seminar on linear algebra is coming up on 14th November 2019, (8 PM IST).

## Linear Algebra total recall (Open Seminar)

Open Seminar on linear algebra. A review of all major ideas. Even if you have little or no knowledge about Linear Algebra, you may join. Register now.

## 4 questions from Sylow’s theorem: Qn 4

I have prepared some common questions ok application of Sylow’s theorem with higher difficulty level. It is most propably cover all possible combination.

## 4 questions from Sylow’s theorem: Qn 3

I have prepared some common questions ok application of Sylow’s theorem with higher difficulty level. It is most propably cover all possible combination.

## 4 questions from Sylow’s theorem: Qn 2

I have prepared some common questions ok application of Sylow’s theorem with higher difficulty level. It is most propably cover all possible combination.

## 4 questions from Sylow’s theorem: Qn 1

I have prepared some common questions ok application of Sylow’s theorem with higher difficulty level. It is most propably cover all possible combination.