agsfaltex wrote:

I came to this question on my second GMAT Prep exam. Please help.

Tanya prepared 4 different letters to be prepared for 4 different addresses. For each letter, she prepared an envolope with the correct address. If the 4 letters are to be put in the 4 envolopes at random, what is the probability that only 1 letter will be put into the envolope with the correct address.

A (1/24)

B (1/8)

C (1/4)

D (1/3)

E (3/8)

Total ways = 4! = 24

Only one letter with correct addr. and three letters with wrong addr.

=> Assume you already put the letter with the correct addr.. then it becomes 3 letters with wrong addr..

Total ways of putting 3 letters with wrong addr. = 2*1*1 =2

Thus, 2*4 = 8 (since there are 4 letters)

8/24 = 1/3

D it is.

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