Only 2 ways to 4-color regular tetrahedron, looking for simple answer A regular tetrahedron is a triangular pyramid whose faces are all equilateral triangles. How many distinguishable ways can we paint the four faces of a regular tetrahedron with red, blue, green, and orange paint such that no two faces are the same color?
I searched online and the answer given was $2$, but all the explanations went over my head. Can someone give me a simple answer (understandable to a primary schooler) as to why that's the case, and avoids discussion of Burnside, orbit-stabilizer, groups, etc.
 A: Suppose that you are looking down at the pyramid from above (the apex) such that the face painted orange is facing downwards and you cannot see it. Any painting can be brought to this configuration by rotating the pyramid appropriately.
How many ways is it possible to paint the remaining three faces that you can see such that none of the configuration can be brought to another by rotating? The answer is $2$ -- the colours red, blue, and green are painted either clockwise or counter-clockwise in that order.
Burnside's Lemma essentially provides a way to account for this symmetry (interconversion via rotation) that we are trying to eliminate, and only really comes to use when you have more than three upward faces.
A: Put the tetrahedron on the table onto its red face. Now rotate it (without lifting it off the table) so that the blue face is facing away from you, the other two faces towards you. Those two faces must be green and orange.
You now have only two possibilities: Green on the left and orange on the right, or orange on the left green on the right.
A: Every face of the tetrahedron shares exactly one edge with every other face, and every triplet of faces share exactly one vertex
This means every colour is adjacent to every other colour exactly oncez and every triplet of colours occurs exactly once (at distinct vertices).
Pick any colour triplet or equivalently its vertex. You can arrange the colours in order around it in two unique ways. The fourth colour is on the opposite face regardless of which order.
Since this choice was arbitrary, it doesn't matter which triplet you choose, there are only ever two ways to arrange the colours, and only two ways to colour the tetrahedron.
