Probability prob

The letter which will be put in correct envelope we can select in 4C1=4 ways

among the rest 3 letters 1 letters can be put into wrong envelope in 2C1=2 ways

the rest of the two letters can be put in wrong envelope in 1 way

so total number of ways in which only 1 letter will be put into the envelope with correct address=4*2
and total number of ways of putting 4 letters in 4 envelope is 4!

hence the probability that only 1 letter will be put into the envelope with correct address=4*2/4!=1/3
Ans option D

If we call the letters A, B, C and D, let's say that the correct envolope placement is represented by the right order. so if all letters are placed correctly, we'll have ABCD. However, if only A is placed correctly we'll have one of the following 2 possibilities: ACDB or ADBC

In fact, for each correctly placed letter, there are only 2 ways to incorrectly place the other three. For instance if only letter B were placed correctly we would have the following 2 possibilities: DBAC or CBDA.

Because this operation can be repeated with anyone of the 4 letters, there are a total of 2*4=8 ways to have only 1 letter placed correctly.

The total number of ways possible of arranging the 4 letters is simply 4!=24, so there are 24 ways to arrange the letters, but 8 of these ways will have 1 correct placement and 3 incorrect placements. The probability therefore that there is only1 letter in the correct place is 8/24 = 1/3

TAKE-AWAY: When a combinatorics question asks you to deal with a small set of possibilities, if you're not sure which formula to use or how to use it, you can actually map out possibilities.

If you have trouble understanding this, you can watch a video solution at GMATPrep question 1071

-Patrick