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2025 AMC 10B Problem 11

Problem 11 of 25IntermediateCounting & Probability

On Monday, 66 students went to the tutoring center at the same time, and each one was randomly assigned to one of the 66 tutors on duty. On Tuesday, the same 66 students showed up, the same 66 tutors were on duty, and the students were again randomly assigned to the tutors. What is the probability that exactly 22 students met with the same tutor both Monday and Tuesday?

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Solution

Each day’s assignment is a permutation of the 66 students among the 66 tutors. Comparing the two days, the number who keep the same tutor is the number of fixed points of τ=πTue1πMon,\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}}, itself a uniformly random permutation of 66 elements. We want exactly 22 fixed points, so choose those 22 in (62)\binom{6}{2} ways and derange the other 4,4, where D4=9.D_4 = 9. The probability is (62)D46!=159720=316.\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}. Thus, B is the correct answer.

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Concepts: derangements · permutations · basic probability

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.