coloring a tetrahedron (abstract algebra) Consider a regular tetrahedron. Each of its faces can either be painted blue or red. Up to rotation how many ways can the tetrahedron be painted?
I am thinking that applying Burnside's counting formula will work, but it has been so long since I have used Burnside's counting formula that I do not quite remember how to use it.
 A: Since a tetrahedron has only $4$ faces, and you have only two colors of paint. you can just as easily count them by hand, and in fact there are just $5$ distinguishable ways to paint the faces. However, you want to see how to do it using Burnside's formula.
Burnside's formula says, count the number of invariant colorings for each rotation, add up those numbers, and divide by the number of rotations. In other words, compute the average number of invariant colorings for a random rotation.
The tetrahedron has $12$ rotations: you can rotate it $120^o$ either way about an axis passing through a vertex and the center of the opposite face, $8$ of those; or $180^o$ about an axis through two opposite edges, $3$ of those; or the $0^o$ identity rotation. Take them case by case.
A $120^o$ rotation about an axis through a vertex $v$: for the coloring to be invariant under this rotation, the three faces adjacent to $v$ have to be the same color; in other words the faces are partitioned into $2$ "orbits", one orbit consisting of the three faces adjacent to $v$, the other orbit containing the remaining face. The number of invariant colorings is $2*2=4$. (If your palette contained $c$ colors, the number of invariant colors would be $c^2$.) $8*4=32$.
A $180^o$ rotation about an exis through two opposite edges, $e$ and $f$: as it happens, in this case too there are $2$ orbits; the two faces meeting at $e$ belong to one orbit, so they have to be the same color; likewise the two faces meeting at $f$. Again, there are $4$ invariant colorings for each of these rotations. $3*4=12$.
The identity rotation: each face is a separate orbit, all $2^4=16$ colorings are invariant, $1*16=16$.
The final answer to your question is $(8*4+3*4+1*16)/12=60/12=5$. For coloring the faces of a regular tetrahedron with $c$ colors, the number of distinguishable patterns is $\dfrac{11c^2+c^4}{12}$.
Exercise: The number of distinguishable ways to color the faces of a cube red, white, or blue is

$57$

A: A case analysis shows that the only thing that distinguishes colourings is the number of faces of each colour. Therefore, the answer is $5$.
A: For the general case, we can give an elementary combinatorial solution.
General Case: We have $n$ different colors. The vertices of a regular tetrahedron will be painted using these colors. The colorings obtained from one another as a result of rotating the tetrahedron are considered identical. The number of different colorings is $$ \dfrac{n^4 + 11n^2}{12} .$$
Proof: If we use exactly $4$ colors, there are $2$ different colorings. Because, we apply one of the colors (only $1$ way) to the base. With the remaining $3$ colors, we can paint the faces with $(3-1)!=2$ ways that using circular permutation. The selection number of $4$ out of $n$ colors is $\dbinom{n}{4}$. With the multiplication principle, we get $2\cdot \dbinom{n}{4}$ colorings. Similarly,  if we use exactly $3$ colors, there are $\dbinom{3}{1}\cdot \dbinom{n}{3}$ colorings. If we use exactly $2$ colors, there are $3\cdot \dbinom{n}{2}$ colorings. If we use $1$ color, there are $n$ colorings. In total, we find the number of different colorings is
$$ 2\cdot \dbinom{n}{4} + 3\cdot \dbinom{n}{3} + 3\cdot \dbinom{n}{2} + n = \color{red}{\dfrac{n^4 + 11n^2}{12}} .$$
Especially for $n=2$, we get $\dfrac{2^4 + 11\cdot 2^2}{12} = \dfrac{60}{12} = 5$.
Note: Moreover we can see that, a part of our proof is that it solves the problem. If we use exactly $2$ colors, there are $3\cdot \dbinom{2}{2}$ colorings. If we use $1$ color, there are $2$ colorings. In total, we get $3 + 2 = 5$ different colorings.
