2007 AMC 12B Problem 16

The rotation group of the tetrahedron has $12$ elements: the identity, $8$ rotations of order $3$ about a vertex-face axis, and $3$ rotations of order $2$ about an edge-midpoint axis. The identity fixes all $3^4=81$ colorings. Each vertex rotation fixes one face and cycles the other three, so it fixes $3^2=9$ colorings; likewise each edge rotation swaps two pairs of faces and fixes $3^2=9.$ By Burnside’s lemma the number of distinguishable colorings is $$\dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15.$$ Thus, the correct answer is A.