Possible sides of and octahedron What number of unique patterns can be made if all sides of an equilateral octahedron is blue or green? How do you solve such a problem?
I have only tried to solve this by a hands-on approach, i.e. actually painting the sides of an octahedron and making notes. I would like to find a mathematical approach.
Background: I lay in my bed, came up with this problem and now I am trying to understand how to solve it with the help of mathematics. That's all the background there is. I have no tools. Please don't kill my curiosity for math as an adult. Teachers already did a great job when I was a student.
 A: Let's agree to call two colorings equivalent if they can be gotten from each other by rotating the octahedron (we could also allow reflections, but those involve breaking the octahedron to atoms, so I'm not).
There are $24$ rotational symmetries: you can rotate a given vertex to any of the $6$ positions. Also, leaving a given vertex were it is, we can still rotate the octahedron by any multiple of $90^\circ$ about the axis formed by that vertex and the one opposite.
$24$ rotations fall into following types (called conjugacy classes):


*

*the identity rotation (= do nothing).

*$3$ rotations by $180^\circ$ about one of $3$ axes joining a pair of opposite vertices, such rotations interchange $4$ pairs of faces with each other.

*$6$ rotations by $90^\circ$ about the same axes. Any one of these permutes
two sets of $4$ faces cyclically amongst themselves.

*$8$ rotations by $120^\circ$ about an axis joining the centroids of two opposite faces ($4$ choices for the two faces, two choices for the direction of rotation). Any one of these maps the face intersected by the axis to itself, and cyclically permutes the remaining six faces in two sets of three.

*$6$ rotations by $180^\circ$ about the axis gotten by joining the midpoints of two opposite edges of the octahedron. These too swap the faces in four pairs.


To use Burnside's lemma we consider all possible colorings, and count the number of colorings fixed by that rotation. Observe that a rotation fixes a coloring if and only if the faces mapped to each other by that rotations share the same color.


*

*The identity rotation fixes all $2^8=256$ colorings.

*Rotations by $180^\circ$ acted on the faces as four pairs. We can select either color to each pair, so there are $2^4=16$ colorings fixed by any of these.

*Here the two sets of four faces must both be unicolored, so the $90^\circ$ rotations fix only $2^2=4$ colorings.

*For the $120^\circ$ rotations we can make four independent coloring choices: one for both fixed faces, and one for the two triplets. A total of $2^4=16$ colorings are fixed by these rotations.

*This is the same as in case 2. above.


Burnside's lemma then tells us that the number of indistuinghisable colorings is
$$
\frac1{24}(256+3\cdot16+6\cdot4+8\cdot16+6\cdot16)=23.
$$
A: Coloring the faces of an octahedron is the same as coloring the corners of a cube; the latter is probably easier to picture.
No blue corners is one configuration (call it $0.1$).  One blue corner is one configuration $(1.1)$.  Two blue corners: either the two are on the same edge $(2.1)$, or they're opposite corners on the same face $(2.2)$, or they're diametrically opposite $(2.3)$.  Three blue corners: either all three are on a single face $(3.1)$, or two are on the same edge and the third is on the opposite edge $(3.2)$, or no two are on the same edge $(3.3)$.  That's $8$ configurations so far.  There are $8$ more that come from swapping blues and greens to yield from five to eight blue corners.
Finally, four blue corners.  All four can be on the same face $(4.1)$.  Three can be on the same face, with the fourth on the opposite face; if those three make an L-shape on the front face, the fourth can be in four different positions on the back face $(4.2-4.5)$.  Two can be on one edge with two on the opposite edge $(4.6)$.  Or no two can share an edge $(4.7)$.  Putting these together:
$$\begin{array}{ll} \text{# blue} & \text{# colorings} \\
\hline
\;0 & \;1 \\
\;1 & \;1 \\
\;2 & \;3 \\
\;3 & \;3 \\
\;4 & \;7 \\
\;5 & \;3 \\
\;6 & \;3 \\
\;7 & \;1 \\
\;8 & \;1 \\
\hline \text{total} & 23
\\ \end{array}$$
If you do not count reflections as distinct, then two of the four ways of making an L-shape on the front face and placing the fourth blue corner on the back face are equivalent.  The total number of configurations would then be $22$.
