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:

Seat the first student anywhere. The second student lands next to them with probability $\tfrac{2}{5},$ since two of the remaining five chairs are adjacent. Now the teachers fill two of the four leftover chairs. Of the $\binom{4}{2} = 6$ ways to do that, exactly $3$ are adjacent pairs. So the probability is $\tfrac{2}{5} \cdot \tfrac{3}{6} = \tfrac{1}{5}.$ Therefore, the answer is B.

2025 AMC 10A Problem 14

Solution:

Problem 14 in Other Years