Four letters, none in the right envelope

Question

Four letters are meant for four addressed envelopes, one letter per envelope. A careless clerk stuffs them at random.

In how many ways can all four letters go into wrong envelopes (none in its correct one)?

Accepted Answer

This asks for a derangement of 4 items, D4. Using Dn = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdotsight), for n = 4: D4 = 24\left(1 - 1 + 	frac{1}{2} - 	frac{1}{6} + 	frac{1}{24}ight) = 12 - 4 + 1 = 9. The recurrence confirms it: D4 = 3(D3 + D2) = 3(2 + 1) = 9. Takeaway. Letters-in-wrong-envelopes is the classic derangement. Here D4 = 9.

Four letters, none in the right envelope

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