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.
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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