I don't think this is solved by Burnside's Lemma since there is a condition that each side is painted a different colour. The question is as follows.

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?

The answer given in an old thread on a different site is $6!$ for the first question, and $n(n-1)(n-2)(n-3)(n-4)(n-5)$ for the second question. However, this doesn't actually hold up because a few of the paintings are isomorphic. The original thread assumes we can somehow tell the difference between two paintings which actually look identical if you rotate the cube, which I don't think is what the question intended.

The answer I got for the first question is $4! + 4 = 28$. But this was just through a case-bash, and I'm not sure whether it's correct or whether it generalizes.


A cube can be rotated into $6 \times 4 = 24$ configurations (i.e. the red face can be any one 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 $6!/24 = 30$.


Case $n=6$: Colour one side with the ugliest colour, and put the cube on a table ugly side down. There are $5$ choices for the colour on top. For each of these choices, colour the side facing you with the nicest remaining colour. The last three sides can be coloured in $3!$ ways, so the number of colourings is $(5)(3!)$.

Case $n>6$: First choose the colours, then use them. The number of colourings is $$\binom{n}{6}(5)(3!).$$

  • $\begingroup$ I don't quite get the idea for case n=6. Why do you need to apply ugliest and nicest terms? For the first colour choosing (i.e. ugliest one), there are 6 ones to choose, why not 6? $\endgroup$ – herohuyongtao Jan 20 '14 at 6:24
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    $\begingroup$ Two colourings are considered to be the same if one can be made identical to the ither by a rotation of the cube. The "ugliest side down" is a colourful way of showing how to make sure that we are only counting up to equivalence. Alternately, we put the cube colour 1 down. Then the colour of the top face is any of $5$. To make sure we don't double-count colourings equivalent by rotation, we rotate the cube so that the "smallest" remaining colour is facing us. If we do the two things (colour 1 down, smallest side colour facing us) then we have made sure that any differences (Cont) $\endgroup$ – André Nicolas Jan 20 '14 at 6:36
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    $\begingroup$ (Cont) in the colouring of the remaining side faces cannot be gotten rid of by a rotation. $\endgroup$ – André Nicolas Jan 20 '14 at 6:37

protected by Zev Chonoles Feb 24 '16 at 6:42

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