2012 AMC 10A Problem 23

Picture the six people as points and each friendship as a segment between two points. Every person has the same number $k$ of friends. "Some" friendships means $k \ge 1$; "not all" rules out everyone being friends with everyone, so $k \le 4$.

Complement symmetry: if we swap every friendship for a non-friendship and vice versa, a configuration in which everyone has $k$ friends becomes one in which everyone has $5 - k$ friends, and swapping back undoes it. So there are as many $k = 3$ configurations as $k = 2$, and as many $k = 4$ as $k = 1$. Only $k = 1$ and $k = 2$ need to be counted.

$k = 1$: the six people split into three friend pairs. Adam's friend can be any of $5$ people; the first person not yet paired has $3$ choices; the last pair is forced. That is $5 \cdot 3 = 15$.

$k = 2$: each person's two friendships link into closed loops that cover all six people, and a loop needs at least three people, so the loops are either one loop of six or two loops of three.

• One loop of six: seat six people around a circle, $\frac{5!}{2} = 60$ ways, because rotating or reflecting the circle gives the same friendships.
• Two loops of three: choose Adam's two loop-mates in $\binom{5}{2} = 10$ ways; the other three form the second loop.

So $k = 2$ gives $70$, hence $k = 3$ gives $70$ and $k = 4$ gives $15$. Total: $15 + 70 + 70 + 15 = 170$.

The answer is $\boxed{\textbf{(B)}\ 170}$.

"Everyone has the same number of friends" describes a regular graph, and two structural facts make the count tractable: taking the complement swaps degree $k$ with degree $5 - k$, and a graph where every vertex has degree $2$ is a disjoint union of cycles. Whenever a condition is symmetric under exchanging "yes" and "no," count only the small half and double it.

Check the $k = 3$ count directly. A configuration where everyone has three friends on six people is either "two groups of three, everyone friends with all of the other group" (choose the groups: $\binom{6}{3}/2 = 10$) or "two triangles with matching corners joined" (a triangular prism: $6!/12 = 60$ labelings). Total $70$, agreeing with the complement argument.

• Forgetting that the two-friends case includes two separate triangles as well as one six-person loop, or overcounting loops by ignoring rotations and reflections.
• Including $k = 0$ or $k = 5$ (one configuration each), which the phrase "some, but not all" excludes.
• Counting the six-person loops as $6! = 720$ or $5! = 120$ instead of $60$.