Task: In how many ways can we color the edges of a tetrahedron with black and red colors?
I have a problem because I computed the cycle index, considering the rotations of the tetrahedron based on its edges, and obtained: $Z(G,x_1,x_2,x_3)=\frac{1}{12}(x_1^6+8x_3^2+3x_1^2x_2^2)$ or $Z(G,2,2,2)=12$, so the answer is 12.
Then I thought like this: if I label the tetrahedron based on its edges and observe its neighbors (adjacent edges), I get exactly an octahedron, which has a cycle index: $Z(G,x_1,x_2,x_3,x_4)=\frac{1}{24}(x_1^6+6x_1^2x_4+3x_1^2x_2^2+6x_2^3+8x_3^2)$ or $Z(G,2,2,2,2)=10\neq 12$.
So what is correct and where is the mistake?
12 is the right answer. The edges of tetrahedron "transform" to vertices of octahedron, which builds bijection. However, bijection is simply not enough. Take a look at those pictures:Tetrahedrons Octahedrons
There are only 12 different ways to color vertices of tetrahedron. Bijection transforms them into the following octahedrons. However, there are 2 pairs of the same graph for octahedrons.