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Question: A cube is painted with six different colours, with each face painted a different colour. In how many different ways can this be done? Note that two colourings are regarded as the same if one can be rotated onto another.

This is from my school's summer homework as well.

I have no clue on how to attempt this question, I only know that there are $6!$ ways to colour a cube this way but I do not know how to get rid of the colourings of which when you rotate the cube it looks like another colouring. $6!$ is $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ which is $720$. This is because if you start with $6$ colours, you have $6$ choices for the first face, $5$ for the second face and so on down to only $1$ colour for the last remaining face.

How can I solve this problem? Because I have no idea on how to make any more progress.

MathsFreak
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3 Answers3

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In general, you have to use Burnside's lemma for such problems. But when there are 6 diff colors, things get much easier.

We can count by construction while keeping rotational equivalency in mind, while we construct.

Let's denote colors as color 1 thru 6. Color 1 has to go somewhere. So put it somewhere.

  1. Opposite of color 1, there are 5 diff choices.

  2. There are 4 colors to fill up the 4 faces adjacent to color 1. But you can rotate around the axis going thru color 1 and its opposite face. So there are only 3! choices.

Answer 5*3!

Shang Zhang
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Please see the following link - it's have similarity regarding the rotation issue. About the 6! combination - it is correct.

Itzik Chaimov
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If the total number of colourings of a cube that can be rotated is $N$, the total number of colourings of a cube that is fixed in space is $Nk$ where $k$ is the number of ways to rotate a cube. For each combination, we pick a colouring and a rotation of that colouring. (We don't need to worry about double counting here. All 6 colours are different, so every time we rotate the cube we get a colouring that looks different.) You've already computed $Nk=6!=720$. We want to find $N$, so now we just need to find $k$ and divide.

Computing $k$ is not too hard. Call a face of the cube face $A$. It can point in 6 different directions. Call an adjacent face of the cube face $B$. It can point in any of the 4 directions orthogonal to the direction face $A$ is pointing. That means there are $k=6\times 4 = 24$ rotations.

So $N = 720 / 24 = 30$

pb1729
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