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?

$\dfrac{1}{16}$

$\dfrac{3}{16}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

Solution:

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.

2025 AMC 10B Problem 11

Solution:

Problem 11 in Other Years