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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?

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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

$$ \frac{1\cdot 3^4+2\cdot 3+1\cdot 3^2}{4}=24. $$

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  • $\begingroup$ Yes, you are right. The amended calculation is the one that I was supposed to find. Thank you very much ! $\endgroup$
    – user403023
    Jun 13 '17 at 14:48

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