How many regular tetrahedrons of the same size can we distinguish using 4 different colours to paint all their faces? Each face can only be painted with one colour.
My solution:
- Using a single colour to colour the whole tetrahedron, we can form 4 distinct tetrahedrons, each with a different colour;
- Using two colours to colour the tetrahedron, we can choose $\binom{4}{2}$ colour combinations. In addition, we can colour three faces with one colour and one face with the other colour, and vice versa, or two faces with each colour. totalling $6 \times 3=18$ distinct tetrahedrons;
- Using three colours to colour the tetrahedron, we can choose $\binom{4}{3}$ colour combinations. In addition, we can choose $3$ options for the colour that will colour two faces, while the other two colours will be used to colour a single face each. used to colour a single face each, generating in this case, of $4\times3=12$ possibilities;
- Using the four colours, each colour will colour one face, and it will only be possible to distinguish $2$ tetrahedrons by reversing the colours of any two faces. Any two faces.
The sum of the cases results in $36$.
As the tetrahedron has 4 faces and equal sides, I was wondering if this problem could be interpreted as: how many different ways can I arrange 1, 2, 3 or 4 people on a round table with positions? If so (or something involving circular permutation and/or circular permutation with repetition) I'd like a hint on how to proceed with the solution.