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.