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Suppose we want to paint the faces of a tetrahedron using $4$ different colors, assuming that we allow different faces to be painted with the same color.

By not taking the symmetries of tetrahedron into account, there are $|X|=4^4=256$ ways. Now, when its symmetries are introduced, from Burnside's orbit-counting theorem: $$ r=\frac{1}{|G|}\sum_{g \in G}|X_g| $$ where $G$ is the symmetry group of the tetrahedron (considering only orientation preserving symmetries), $X_g$ are the elements fixed by $g$ and $r$ is the number of orbits of $X$ under $G$'s action. Therefore, we need to keep track of how many elements are fixed by every $g \in G$.

$\bullet \space $The identity element keeps everything unchanged, so $X_e=256$

$\bullet \space $ Let $\rho^j_i$ denote the rotations about the vertex $i$ by $j$ degrees. For the element to stay fixed, the adjacent faces to this vertex must be of the same color. So $$|X_{\rho^{120}_1}|=|X_{\rho_1^{240}}|=\dots=|X_{\rho^{240}_4}|=4 \cdot4=16$$

$\bullet \space $ There are $3$ more symmetries left to examine, which are the $180^o$ rotations, $m_1,m_2, m_3$. Elements stay fixed only if they have two pairs of similarly colored adjacent faces. Thus $$ |X_{m_1}|=|X_{m_2}|=|X_{m_3}|=4 \cdot 4=16 $$ Taking all this into account, we yield: $$ r=\frac{1}{12}(256+8 \cdot 16 + 3 \cdot 16)=\frac{432}{12}=36 $$

Therefore, there are $36$ different ways to color the faces of a tetrahedron using $4$ colors.

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  • $\begingroup$ You have used only orientation preserving symmetries. The tetrahedron also has orientation reversing symmetries. $\endgroup$ – Lee Mosher Jan 31 '19 at 19:28
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    $\begingroup$ @LeeMosher I'm assuming it's a real-world 3D tetrahedron, so $G \cong A_4 \leq S_4$. $\endgroup$ – Andrew Jan 31 '19 at 19:31
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    $\begingroup$ Well, even orientation reversing symmetries are regarded as being "real", i.e. bilateral symmetry. Regardless, to be mathematically precise you can say that you are considering only orientation preserving symmetries. You can edit your question to say so. $\endgroup$ – Lee Mosher Jan 31 '19 at 19:34
  • $\begingroup$ @LeeMosher Given this consideration, is my approach correct? $\endgroup$ – Andrew Jan 31 '19 at 20:07
  • $\begingroup$ Looks correct to me! $\endgroup$ – Mike Earnest Jan 31 '19 at 20:15
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As stated in the comments, the calculation appears to be correct for the orientation-preserving symmetries of the tetrahedron.

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