= Solution
{c}
For a <finite group> action, <Burnside's lemma> gives
$$
|\Omega/G|=\frac1{|G|}\sum_{g\in G}|\operatorname{Fix}(g)|.
$$
Pair each vertex with its opposite face. This correspondence is equivariant under every tetrahedral symmetry. Each pair has six possible combined color states: three face colors and two vertex colors. Thus we are coloring four positions with six symbols; the two kinds of coloring must be counted together.
The group of rotations is the <alternating group> $A_4$. The identity has four cycles and fixes $6^4$ colorings. Its three half-turns have cycle type $(2,2)$ and each fix $6^2$. Its eight third-turns have cycle type $(3,1)$ and each also fix $6^2$. Consequently the rotation count is
$$
\boxed{\frac{6^4+3\cdot6^2+8\cdot6^2}{12}=141.}
$$
Allowing mirror equivalence gives the full tetrahedral symmetry group $S_4$. Now all permutations of the four paired positions are allowed, so a coloring is determined by a <multiset> of four symbols chosen from six. <stars and bars> gives
$$
\boxed{\binom{6+4-1}{4}=\binom94=126.}
$$
For an independent Burnside check, the five cycle types of $S_4$ give numerator $6^4+6\cdot6^3+3\cdot6^2+8\cdot6^2+6\cdot6=3024$, whose quotient by 24 is 126. This is <paired face and vertex colorings of a tetrahedron>.
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