2012 AMC 12A Problem 19

Model people as vertices of a graph, with edges for friendships. Everyone has the same degree $n$ with $1 \le n \le 4.$ The cases $n$ and $6 - 1 - n$ are complementary graphs, so $n = 1$ pairs with $n = 4$ and $n = 2$ with $n = 3.$ For $n = 1$ the graph is a perfect matching: $5 \cdot 3 = 15$ ways. Thus $n = 4$ also gives $15.$ For $n = 2$ the graph is a union of cycles: either two triangles $\left(\binom{5}{2} = 10\right)$ or one hexagon $\left(\dfrac{6!}{12} = 60\right),$ totaling $70.$ Thus $n = 3$ also gives $70.$ The total is $15 + 15 + 70 + 70 = 170.$ Thus, the correct answer is B.