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?

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{2}{9}$

$\dfrac{3}{13}$

$\dfrac{1}{4}$

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.

2025 AMC 12A Problem 6

Solution:

Problem 6 in Other Years