2020 AMC 10B Problem 17

Number the people $0, 1, 2, \ldots, 9$ around the circle. A person can pair with a neighbor (like $3$ and $4$) or with the person directly across (like $3$ and $8$, numbers that differ by $5$). There are $5$ across pairs: $\{0,5\}, \{1,6\}, \{2,7\}, \{3,8\}, \{4,9\}$.

A useful fact. Suppose some people are already paired. The rest sit in stretches of people standing next to each other, and they must pair with neighbors inside their stretch. A stretch like $1, 2, 3, 4$ can be paired in only one way: $\{1,2\}, \{3,4\}$. A stretch with an odd number of people cannot be paired at all.

Now split into cases by how many across pairs are used.

1. No across pairs. Everyone pairs with a neighbor around the full circle: either $\{0,1\}, \{2,3\}, \ldots, \{8,9\}$ or $\{1,2\}, \{3,4\}, \ldots, \{9,0\}$. That is $2$ ways.
2. One across pair. Say $\{0,5\}$. The others form two stretches, $1,2,3,4$ and $6,7,8,9$, each with $4$ people, so each pairs up in one way. There are $5$ choices of across pair, so $5$ ways.
3. Two across pairs. Try $\{0,5\}$ and $\{1,6\}$: the stretches left are $2,3,4$ and $7,8,9$, with $3$ people each. Try $\{0,5\}$ and $\{2,7\}$: the stretches are $1$, then $3,4$, then $6$, then $8,9$, and the single person $1$ cannot pair. Every choice of two looks like one of these, so $0$ ways.
4. Three across pairs. Now look at the two across pairs not used; those $4$ people must pair with neighbors. If the unused pairs are $\{0,5\}$ and $\{1,6\}$, then $\{0,1\}$ and $\{5,6\}$ work. If they are not next to each other, like $\{0,5\}$ and $\{2,7\}$, then person $0$ has no free neighbor. There are $5$ ways to choose two neighboring across pairs ($\{0,5\}\&\{1,6\}$, $\{1,6\}\&\{2,7\}$, $\{2,7\}\&\{3,8\}$, $\{3,8\}\&\{4,9\}$, $\{4,9\}\&\{0,5\}$), so $5$ ways.
5. Four across pairs. The two people left stand directly across from each other, and they are not neighbors. $0$ ways.
6. Five across pairs. $1$ way.

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

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

Choosing the special across pairs first leaves people standing in rows, and a row pairs up in exactly one way when it has an even number of people and in no way when it is odd. Checking that even-or-odd rule for each case is what knocks out the cases with $2$ and $4$ across pairs, which students often count by mistake.

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 $\boxed{\textbf{(C)}\ 13}$.