counting problem

Total # of arrangements possible to put 4 letters in 4 envelopes = 4! = 24

If exactly one letter is put in correct envelope, those cases are:

A C D B
A D B C
D B A C
C B D A
B D C A
D A C B
B C A D
C A B D

Total 8 such cases when only 1 letter is put in the correct envelope.

Reqd prob = 8/24 = 1/3

Just some further analysis of this problem:

Had it been to find the # of arrangements with exactly 2 letters put in the correct envelope, that would equal 4C2 = 6 ways

A B D C
A C B D
A D C B
D B C A
C B A D
B A C D

# ways where all letters in their correct envelopes: 1 (A B C D)

# of ways where no letter in correct envelope = 24 - (All correct) - (1 correct) - (2 correct)
= 24 - 1 - 8 - 6 = 9 ways

Another way to find # ways where NO letter is put in its correct envelope = 4! * [1/2! - 1/3! + 1/4!] = 9

General formula: n! * [1/2! - 1/3! + 1/4! - .... 1/n!]