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A small class of nine boys are to change their seating arrangement by drawing their new seat numbers from a box. After the seat change, what is the probability that there is only one pair of boys who have switched seats with each other and only three boys who have unchanged seats?

How do you solve this? I've thought of $\frac{(9C2)(7C3)(4!)}{9!}$ to give the answer, but then I realize the $4!$ includes arrangements for the 4 boys switching seats with each other and having unchanged seats. How do I account for these conditions in this problem?

SolvingTrainee
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  • It's okay. In the arrangement of the $4$ boys it is one of the case (included in $4!$) that the $4$ boys have unchanged seats. – SarGe May 15 '20 at 04:07
  • Once you've chosen the $4$ boys who neither stay still nor switch seats, they have to permute in a $4$-cycle. For each choice of those $4$ boys, there are $6$ possible $4$-cycles, so replace $4!$ with $3!$ and you have your answer. – Robert Shore May 15 '20 at 04:47

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