# Colouring faces of acube

1. May 11, 2013

### Saitama

1. The problem statement, all variables and given/known data
I was asked this question by my friend, unable to do it I did a google search. Here's the question:
http://math.stackexchange.com/questions/64857/painting-the-faces-of-a-cube-with-distinct-colours

I am quoting the question and the answer posted at stackexchange:
2. Relevant equations

3. The attempt at a solution
I need some help in understanding the solution. The first post (answer) in that link finds out the number of configurations. Doesn't these configurations depend on the colour of the face which is just opposite to the red coloured face? I can't even understand the step the poster has done next.

Any help is appreciated. Thanks!

Last edited: May 11, 2013
2. May 11, 2013

### tiny-tim

Hi Pranav-Arora!
No, at least not the number of different configurations that the question asks for.

The first post finds the number of same configurations, ie the number that can be the same as each other (24) …

there are 6! ways of colouring 6 objects,

but on a cube each way is "duplicated" 24 times, so there must be 6!/24 cubely-different ways.

3. May 11, 2013

### Saitama

I still don't get it, for example if I fix red and blue colour on the top and the bottom face respectively, I have 24 ways to colour the side walls but they are all same. Similarly I can go about fixing two other colours on both the sides i.e there are 6C2=15 ways to select two colours and 15 different ways to colour the six walls. But as stated above, the answer is 30, where am I wrong?

4. May 11, 2013

### sjb-2812

5. May 11, 2013

### tiny-tim

Hi Pranav-Arora!
No, do it this way:

first, colour the whole cube

(these colours will remain fixed … all we now do is rotate the cube)

red can be in 6 positions

now there are only 4 positions for the rest of the cube (rotating about the axis through the red face) …

total 24 positions for the same configuration

6. May 11, 2013

### Saitama

I am stuck at this one.

One of the face is coloured red. We are left with 5 colours now. We paint the side walls with 4 out of these 5 colours, let the colours on the side walls be green, blue, yellow and black. There are four same configurations. Let the 6th colour be white. Interchanging green with white, I get more 4 same configurations. Interchanging more colours, there are 20 same configurations for a single position of red.

I hope you understand what I am trying to convey.

7. May 11, 2013

### tiny-tim

No, you're counting (wrongly) the number of different colourings …

the position of the first colour (say, red) doesn't matter, then you have 5 choices for the opposite colour, the position of the third colour doesn't matter, and there are 6 ways of arrangeing the last three colours

total different ways = 5*6 = 30.​

The stackexchange method was to count the number of orientations for the same colouring, result 24, and then divide 6! by 24 to get 30.

8. May 13, 2013

### Saitama

Thanks a lot tiny-tim, I understand it now.