2016 AMC 10A Problem 18

Opposite faces together use all eight labels, whose sum is $36$, so every face must have sum $18$. Put label $1$ at one vertex, and let the three adjacent labels be $a,b,c$. The three vertices adjacent to those across faces are then forced to be $$\begin{gathered} 17-a-b, \\ \quad 17-a-c, \\ \quad 17-b-c, \end{gathered}$$ and the opposite vertex is $a+b+c-16$. Assume the three neighbor labels are listed in increasing order. Substituting triples from $2,3,\ldots,8$ into the four forced expressions above gives exactly $$\begin{array}{c|c} \{a,b,c\} & \text{four forced labels} \\ \hline \{4,6,8\} & \{7,5,3,2\} \\ \{4,7,8\} & \{6,5,2,3\} \\ \{6,7,8\} & \{4,3,2,5\} \end{array}$$ For each neighbor set, the six orders around the vertex fall into two groups of three under the $120^\circ$ rotations about the diagonal through that vertex. Thus each set gives two non-rotationally equivalent arrangements, for $3\cdot2=6$ arrangements in all. Thus, the correct answer is C.