Five gifts with exactly one going home right

Question

Five friends' gifts are shuffled and returned at random, one to each friend. In how many ways does exactly one friend receive their own gift (and the other four do not)?

Accepted Answer

"Exactly one match" splits into two independent choices. First, choose which single friend is the lucky one who gets their own gift: \binom{5}{1} = 5 ways. Then the remaining 4 friends must all get wrong gifts, so those 4 form a derangement: D4 = 9 ways. (It is important they are deranged, otherwise there could be more than one match.) Multiply: \binom{5}{1} 	imes D4 = 5 	imes 9 = 45. Takeaway. "Exactly m fixed" equals (choose the m matched) times (derange the rest): \binom{n}{m} D{n-m}. Here \binom{5}{1} D4 = 45.

Five gifts with exactly one going home right

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