Exactly one correct envelope

Question

A distracted clerk stuffs four different letters into four pre-addressed envelopes completely at random, one letter per envelope.

What is the probability that exactly one letter ends up in its correct envelope?

Accepted Answer

"Exactly one correct" means: choose which single letter is the lucky match, then make sure the other three are all wrong. First pick the one correctly placed letter: there are \binom{4}{1} = 4 choices. The remaining three letters must go into the remaining three envelopes with none of them correct, that is, a derangement of three items. The number of derangements of three items is D3 = 3!\left(1 - 1 + 	frac12 - 	frac16ight) = 6 \cdot 	frac{2}{6} = 2. So the number of favorable arrangements is 4 	imes 2 = 8, out of 4! = 24 total: P(	ext{exactly one correct}) = \frac{8}{24} = \frac{1}{3}…

Exactly one correct envelope

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