How many ways can a cube be painted with distinct colors?

[Saitama]

Homework Statement

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:

If I had a cube and six colours, and painted each side a different colour, how many (different) ways could I paint the cube? What about if I had n colours instead of 6?

A cube can be rotated into 6×4=24 configurations (i.e. the red face can be anyone of the 6, and then there are 4 ways to rotate it that keep that face red), so the number of different colourings (counting rotations, but not mirror reflections, as the same) is 6!/24=30.

Homework Equations

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!

 

[tiny-tim]

Hi Pranav-Arora!

The first post (answer) in that link finds out the number of configurations.

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.

 

[Saitama]

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

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?

 

[sjb-2812]

See perhaps e.g. http://intermolecular.wordpress.com/2009/11/30/chiral-dice/ or similar

 

[tiny-tim]

Hi Pranav-Arora!

… 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.

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 ​

 

[Saitama]

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

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.

 

[tiny-tim]

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.

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.

 

[Saitama]

Sorry for the late reply.

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.​

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