2025 AMC 12A Problem 6

Attempt Problem 6 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

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6.

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?

16\dfrac{1}{6}

15\dfrac{1}{5}

29\dfrac{2}{9}

313\dfrac{3}{13}

14\dfrac{1}{4}

Answer: B
Concepts:circular arrangementsbasic probability
Difficulty rating: 1350
Solution:

Choosing 22 chairs for the students and 22 for the teachers gives (62)(42)=156=90\binom{6}{2}\binom{4}{2} = 15 \cdot 6 = 90 equally likely outcomes.

A round table has 66 adjacent pairs of chairs. Give the students any adjacent pair; among the remaining 44 chairs there are exactly 33 adjacent pairs for the teachers. That is 63=186 \cdot 3 = 18 favorable outcomes.

The probability is 1890=15.\dfrac{18}{90} = \dfrac{1}{5}.

Thus, the correct answer is B.

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