Burnside's lemma: 30 possible different dice I have been working with Burnside's counting lemma and I came across the problem to show that there are 30 possible different dice.
I have tried working with the 24 rotational symmetries of the cube but can't get to an answer.  Any help please?
 A: The answer "$30$" tells me that the orientation of the number on a face of the die doesn't make a difference. In that case, a clearer way to state the problem would be, you have $6$ different colors of paint, how many distinguishable ways can you paint each face of the cube a different color? With this restriction, there are no invariant colorings for any of the rotations except the identity, so Burnside's lemma gives$$\frac{6!+0+0+\cdots+0}{24}=30.$$Of course there are many other ways to arrive at this answer. For example, we can arbitrarily start by painting the bottom red. There are then $5$ choices of color for the top, and $3!$ cyclic orderings for the other $4$ colors, so again we get the answer $5\cdot3!=30$.
If, instead of coloring the faces of the cube, you number them from $1$ to $6$, with the numbers represented by patterns of spots as on an ordinary gambling cube, and if you take into account the fact that the patterns of two, three, or six spots can be oriented in two different ways, the number of different dice is then $2^3*30=240$.
