University of South Carolina High School Math Contest/1993 Exam/Problem 11
Suppose that 4 cards labeled 1 to 4 are placed randomly into 4 boxes also labeled 1 to 4, one card per box. What is the probability that no card gets placed into a box having the same label as the card?
This is the number of derangements of 4 objects. We can know the formula for derangements or count in one of two ways:
Counting directly: The card labeled 1 has 3 places to go. Without loss of generality we may say that it goes in the place marked 2. Now, if card 2 goes into box 1, we have only one possibility because cards 3 and 4 must be interchanged. Otherwise, there are 2 possibilities for card 2, and each of these leads to one more possible arrangement (if card 2 goes in box 3, card 3 must go in box 4 and card 4 in box 1, while if card 2 goes in box 4, card 4 must go in box 3, and card 3 must go in box 1). This gives us a total of good arrangements. (Equivalently, one could say that the only permutations of 4 objects with no fixed points are those with cycle notation and , of which there are 6 and 3, respectively.) Thus the probability is .
Counting the complement:
Using the formula for derangements, , we can substitute in our values to find that the number of derangements is Therefore, the number of derangements is 9. The probability of getting a derangement is Simplifying, we get .