2025 AMC 10B Problem 11

Question

On Monday,

6

students went to the tutoring center at the same time, and each one was randomly assigned to one of the

6

tutors on duty. On Tuesday, the same

6

students showed up, the same

6

tutors were on duty, and the students were again randomly assigned to the tutors. What is the probability that exactly

2

students met with the same tutor both Monday and Tuesday?

\dfrac{1}{16}

\dfrac{3}{16}

\dfrac{1}{4}

\dfrac{3}{8}

\dfrac{1}{2}

Each day’s assignment is a permutation; comparing the two days is like a single random permutation, and “same tutor” means a fixed point

Choose the

2

matched students in

\binom{6}{2}

ways and derange the other

4

(there are

D_4 = 9

derangements); divide by

6!

Each day’s assignment is a permutation of the

6

students among the

6

tutors. Comparing the two days, the number who keep the same tutor is the number of fixed points of

\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}},

itself a uniformly random permutation of

6

elements. We want exactly

2

fixed points, so choose those

2

in

\binom{6}{2}

ways and derange the other

4,

where

D_4 = 9.

The probability is

\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}.

Thus, B is the correct answer.

Problem 11 in other years