2025 AMC 10A Problem 14

Number the chairs $1$ to $6$ around the table, so chair $6$ is next to chair $1$.

Whether two people are side by side depends only on which two chairs they take, not on who sits in which one. So think of it this way: the students pick a pair of chairs, then the teachers pick a pair from the chairs left over. Every pair is equally likely at each step.

Step 1: the students.

1. The number of ways to pick $2$ chairs out of $6$ is $\binom{6}{2} = 15$.
2. The side-by-side pairs are $\{1,2\}, \{2,3\}, \{3,4\}, \{4,5\}, \{5,6\}, \{6,1\}$. That is $6$ pairs (do not forget $\{6,1\}$, which wraps around the table).
3. So the chance the students sit together is $\frac{6}{15} = \frac25$.

Step 2: the teachers, once the students are together.

1. Every side-by-side pair looks the same around a round table, so suppose the students are in chairs $1$ and $2$. The chairs left are $3, 4, 5, 6$.
2. The teachers can pick $\binom{4}{2} = 6$ pairs from these four chairs.
3. Which of those pairs are side by side? $\{3,4\}$, $\{4,5\}$ and $\{5,6\}$. What about $\{6,3\}$? Chairs $6$ and $3$ are not neighbors, because chairs $1$ and $2$ sit between them. The four empty chairs form a row, not a circle.
4. So the chance the teachers sit together is $\frac{3}{6} = \frac12$.

Step 3: combine. Both things must happen, so multiply:
$$\frac25 \cdot \frac12 = \frac15.$$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{5}}$.

Counting pairs of chairs instead of seatings keeps the numbers small, and it is safe because swapping the two students (or the two teachers) never changes who is next to whom. The key moment is Step 2: taking two neighboring chairs out of a circle leaves a row, and a row of $4$ chairs has only $3$ neighboring pairs, while a circle of $4$ would have $4$. Always check whether what is left is still a circle.

Count everything directly instead of multiplying chances.

• All outcomes: the students pick one of $15$ pairs, then the teachers pick one of the $6$ pairs from the four chairs left. That is $15 \cdot 6 = 90$ outcomes, all equally likely.
• Good outcomes: the students pick one of the $6$ side-by-side pairs. For each one, Step 2 above showed the teachers have $3$ side-by-side pairs left. That is $6 \cdot 3 = 18$ good outcomes.

So the probability is $\frac{18}{90} = \frac15$.

Only which pair of chairs each group occupies matters, not who sits where inside a pair. So think of it as the students choosing $2$ of the $6$ chairs, then the teachers choosing $2$ of the remaining $4$; every one of the $\binom{6}{2}\binom{4}{2} = 15 \cdot 6 = 90$ outcomes is equally likely.

Students adjacent. Around a circle of $6$ chairs there are exactly $6$ adjacent pairs, so this has probability $\dfrac{6}{15} = \dfrac{2}{5}$.

Teachers adjacent, given that. Removing two neighbouring chairs from the circle leaves four chairs in a row: for students in chairs $1$ and $2$, the chairs $3, 4, 5, 6$ remain, and chairs $3$ and $6$ are not neighbours, since $1$ and $2$ sit between them. Of the $\binom{4}{2} = 6$ pairs available to the teachers, the adjacent ones are $\{3,4\}, \{4,5\}, \{5,6\}$, so the probability is $\dfrac{3}{6} = \dfrac12$.

Multiplying,
$$P = \frac{2}{5}\cdot\frac{1}{2} = \frac{1}{5}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{5}}$.