How many rotationally distinct coloring of the faces of a pyramid (with square base which we do not color) are there if we use yellow, white and green?
I used Burnside's lemma for this assignment and I got:
$$ \frac{1 \cdot 3^4 + 2 \cdot 3 + 1 \cdot 3^2 + 2 \cdot 3^3 + 2 \cdot 3^2}{8} = \frac{81+6+9+54+18}{8}=21 $$
There is one identity element which leaves all $3^4$ of the elements unchanged.
There is one $90$ degree and one $270$ degree rotation thus there are $3$ possible colors for each.
There is one $180$ degree rotation I have $3 \cdot 3 = 3^2$ possible colors.
I have also facing faces symmetry. There are two kinds of symmetry here: Fix two facing faces and there are $3\cdot 3$ ways of coloring the fixed faces while the other two must have the same color so there are $3$ possible colorings. So therefore there are $2\cdot 3^3$ possible colorings.
The last is diagonal symmetry. I need only to color the two faces in one side of symmetry, so $2\cdot 3^2$.
Is this correct?
Your calculation is correct ... if you're incorporating orientation-reversing symmetries (i.e. reflections), but I am skeptical since the problem says rotationally distinct coverings, which does not seem to allow reflections as symmetries.
The amended calculation would simply be
$$ \frac{1\cdot 3^4+2\cdot 3+1\cdot 3^2}{4}=24. $$