Five students forget their backpacks in class. Their teacher returns the backpacks randomly. What is the probability that exactly two of them get their own backpacks?
I know that there are 5! ways in which the backpacks can be returned to the students and by a simple counting argument I found that there are 20 different ways in which two students get their own backpack returned. And so the answer is a probability of 1/6. My question is what is a formulaic way to come up with 20 different ways in which two students get their own backpack retuned instead of just literally counting the number of different ways?
Count the number of ways a specific pair get their packs back first, say 1 and 2. Then you must identify student 1 with pack 1, student 2 with pack 2, and the rest cannot have their own packs. Thus student 3 gets pack 4 or pack 5. You should see that each of the last two cases leaves only one choice for students 4 and 5, so there are two permutations for 1 and 2 getting their own packs.
So then there would also be two permutations for 1 and 3, two for 1 and 4, and so on. Thus you multiply the two permutations for each specific pair by $\binom{5}{2}=10$ possible pairs getting 20 "correct" permutations in all and a probability of 1/6.