Proper coloring of a cuboid.

Question

Question: The different faces of a cuboid, having the measurements $1 \times 2 \times 2$ cm, is to be painted. Colors at disposal are blue, yellow, and red. In how many ways can this be done, if two colorings are considered as being equivalent when one of them can be obtained from the other one by a rotation of the cuboid?

I was thinking it may be something in this way? Because the rotations should be the same as the ones for a cube? right?

• one identity element which leaves all 3^6 elements of X unchanged
• six 90-degree face rotations, each of which leaves 3^3 of the elements of X unchanged
• three 180-degree face rotations, each of which leaves 3^4 of the elements of X unchanged
• eight 120-degree vertex rotations, each of which leaves 3^2 of the elements of X unchanged
• six 180-degree edge rotations, each of which leaves 3^3 of the elements of X unchanged

Then after putting the numbers inte Burnsdies formula i get the answer 57, someone who can confirm that im doing this the right way?

Answer 1

As discussed in the comments, your solution is correct for a cube. The cuboid has a smaller rotation group: The identity, two rotations through $\frac\pi2$ and one through $\pi$ about the long axis, two rotations through $\pi$ about the axes through the midpoints of the long faces, and two rotations through $\pi$ about the axes through the midpoints of the long edges, for a total of $8$.

• The identity again leaves $3^6$ colourings invariant
• The two rotations through $\frac\pi2$ about the long axis leave $3^3$ colourings invariant.
• The three rotations through $\pi$ about all three axes through the midpoints of faces leave $3^4$ colourings invariant.
• The two rotations through $\pi$ about the axes through the midpoints of the long edges leave $3^3$ colourings invariant.

The total is $3^6+2\cdot3^3+3\cdot3^4+2\cdot3^3=1080$, so there are $\frac{1080}8=135$ inequivalent colourings.