2025 AMC 10A Problem 14

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

Because the two students are interchangeable for the purposes of the question, and so are the two teachers, counting unordered pairs of chairs removes a factor of $2 \cdot 2$ from both the numerator and the denominator. The conditional step is where the circle matters: once an adjacent pair is removed, the survivors form an arc, not a circle, and an arc of $n$ chairs has $n - 1$ adjacent pairs rather than $n$. Keeping track of which figure is still a cycle is the whole content of the problem.

Count directly. There are $6$ adjacent pairs for the students; for each, exactly $3$ of the $6$ adjacent pairs are disjoint from it (the two pairs sharing a chair with it are excluded), so $6 \cdot 3 = 18$ of the $90$ pair-assignments work, giving $18/90 = 1/5$. Equivalently, in terms of seatings, the numerator is $18 \cdot 2 \cdot 2 = 72$ and the denominator $6 \cdot 5 \cdot 4 \cdot 3 = 360$.

• Treating the four chairs remaining after the students sit as if they still formed a circle, which counts a fourth adjacent pair and gives (2/5)(2/3) = 4/15.
• Forgetting that the students' block and the teachers' block are distinguishable, so that the $18$ ordered choices of two disjoint adjacent pairs get halved to $9$.
• Counting only $5$ adjacent pairs among six chairs, as in a row, and losing the pair that wraps around.