On Monday, students went to the tutoring center at the same time, and each one was randomly assigned to one of the tutors on duty. On Tuesday, the same students showed up, the same tutors were on duty, and the students were again randomly assigned to the tutors. What is the probability that exactly students met with the same tutor both Monday and Tuesday?
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Consider an assignment of students to tutors as a permutation of , where the number in the th spot of the permutation corresponds to the student assigned to the th tutor for . Without loss of generality, suppose on Monday the assignment is . Then we are looking for the probability that a permuation on elements has exactly fixed points (spots such that the th element is ).
There are ways to select the students to be fixed points and of the remaning people, none of them can be a fixed point. This corresponds to a derangement of people for a count of ways to assign the remaining people.
Alternatively, if we wish to avoid using derangements, we provide another method to see that there are permutations on elements without fixed points.
Suppose we are arranging so that is not in the th slot for . Then, the first element has options and there are options for the spot that goes to. Regardless of these two choices, there is always exactly one way to fill out the rest of the permutation, for a total of derangements.
The final answer is then:
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