= Paired face and vertex colorings of a tetrahedron
{title2=$N_{\rm rotation}=141,\quad N_{\rm full}=126$}
Pair a <tetrahedron> vertex with its opposite face. This is an equivariant correspondence under all tetrahedral symmetries, so three face colors and two vertex colors become six combined symbols at each of four positions. <Burnside's lemma> on the <alternating group> $A_4$ gives $(6^4+3\cdot6^2+8\cdot6^2)/12=141$. Including reflections gives $S_4$ and reduces the count to <multisets> of four symbols from six, namely $\binom94=126$.
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