2020 AMC 10B Problem 17

Number the people $0$ through $9$ around the circle. The allowed pairs are neighbor pairs $\{i, i+1\}$ and the five "diameters" $\{i, i+5\}$. Classify pairings by how many diameters $k$ they use.

Once the diameters are chosen, the remaining people sit in arcs of consecutive positions and must be paired with neighbors within those arcs. An arc with an even number of people can be paired in exactly one way; an odd arc is impossible.

• $k = 0$: pair everyone with neighbors around the full circle. Either $\{0,1\},\{2,3\},\dots$ or $\{1,2\},\{3,4\},\dots$: $2$ ways.
• $k = 1$: removing one diameter leaves two arcs of $4$ people. Each pairs uniquely: $5$ choices of diameter, $5$ ways.
• $k = 2$: two diameters remove $4$ people at positions $i, j, i+5, j+5$. Whether $i, j$ are adjacent or two apart, some leftover arc has $3$ or $1$ people: $0$ ways.
• $k = 3$: the two unused diameters must leave arcs of even size, which requires their endpoints to be adjacent, e.g. unused diameters $\{0,5\},\{1,6\}$ leave arcs $\{0,1\}$ and $\{5,6\}$. There are $5$ adjacent pairs of diameters: $5$ ways.
• $k = 4$: two leftover people sit directly across, not adjacent: $0$ ways.
• $k = 5$: all diameters, $1$ way.

Total: $2 + 5 + 0 + 5 + 0 + 1 = 13$.

The answer is $\boxed{\textbf{(C)}\ 13}$.

This is counting perfect matchings in a graph (a cycle plus its diameters). Casework on the "special" edges turns the problem into a simple one about paths: a path of even length has a unique matching and an odd path has none. Checking the parity of every leftover arc is what eliminates the $k = 2$ and $k = 4$ cases that most students overcount.

• Assuming any two across pairs can coexist with neighbor pairs; using exactly 2 or 4 across pairs leaves odd arcs and gives zero ways.
• Counting the $k = 3$ case as $\binom{5}{3} = 10$ instead of recognizing only the $5$ choices with adjacent unused diameters work.