Problem Solving

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its complete address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address ?
A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8

With all these letters and envelopes things together I lost my way !!

OA : [spoiler]1/3[/spoiler]

utkalnayak wrote:Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its complete address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address ?
A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8

Let the 4 letters be A, B, C and D.
Total ways to arrange the 4 letters = 4! = 24.
Let the correct ordering of the 4 letters be ABCD.

Write out the ways that ONLY A can be put in the correct position:
ACDB
ADBC
Total ways = 2.

Using the same reasoning, there will be 2 ways that ONLY B can be put in the correct position, 2 ways that ONLY C can be put in the correct position, and 2 ways that ONLY D can be put in the correct position.
Thus, the total number of ways to put EXACTLY 1 letter in the correct position = 2+2+2+2 = 8.

Thus:
P(exactly 1 letter is put in the correct position) = 8/24 = 1/3.

The correct answer is D.

GMATGuruNY wrote: Write out the ways that ONLY A can be put in the correct position:
ACDB
ADBC
Total ways = 2.

Using the same reasoning, there will be 2 ways that ONLY B can be put in the correct position, 2 ways that ONLY C can be put in the correct position, and 2 ways that ONLY D can be put in the correct position.
Thus, the total number of ways to put EXACTLY 1 letter in the correct position = 2+2+2+2 = 8.

Thus:
P(exactly 1 letter is put in the correct position) = 8/24 = 1/3.

The correct answer is D.

In the above case why you think there are only 2 possibilities in each case ?
Should it not be 3! ways ?

utkalnayak wrote:In the above case why you think there are only 2 possibilities in each case ?
Should it not be 3! ways ?

Here's a visual that might help.