Take the 2-minute tour ×  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.  share|improve this question add comment  2 Answers
up vote 8 down vote accepted  A cube ca