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.
