2025 AMC 12A Problem 6

Question

Six chairs are arranged around a round table. Two students and two teachers randomly select four of the chairs to sit in. What is the probability that the two students will sit in two adjacent chairs and the two teachers will also sit in two adjacent chairs?

\dfrac{1}{6}

\dfrac{1}{5}

\dfrac{2}{9}

\dfrac{3}{13}

\dfrac{1}{4}

Answer: B

Concepts:circular arrangements basic probability

Difficulty rating: 1350

Small hint:

Count the ways to give the students one pair of chairs and the teachers another pair

Big hint:

There are

6

adjacent pairs; after fixing the students’ pair, count the adjacent pairs left for the teachers

Solution:

Choosing

2

chairs for the students and

2

for the teachers gives

\binom{6}{2}\binom{4}{2} = 15 \cdot 6 = 90

equally likely outcomes.

A round table has

6

adjacent pairs of chairs. Give the students any adjacent pair; among the remaining

4

chairs there are exactly

3

adjacent pairs for the teachers. That is

6 \cdot 3 = 18

favorable outcomes.

The probability is

\dfrac{18}{90} = \dfrac{1}{5}.

Thus, the correct answer is B.