Query 2

[Nijo]

Tanya had 4 letters to send to 4 different addresses. For each letter she made an envelope with the correct address. If 4 letters are put in 4 envelopes randomly, what is the probability of only 1 letter in the correct envelope
1) 1/24
2) 1/8
3) 1/3
4) 3/8

OA is C
Thanks very much

[GMATGuruNY]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct 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

Approach 1:

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.

Approach 2:

Let the correct order for the letters be A-B-C-D.

Case 1: Only A in the correct envelope
P(A is placed in the CORRECT envelope) = 1/4. (Of the 4 envelopes, 1 is correct.)
P(B is placed in an INCORRECT envelope) = 2/3. (Of the 3 remaining envelopes, only 1 is correct, so 2 are incorrect.)

At this point, B has taken either the envelope for C or the envelope for D.
Thus, of C and D, ONLY ONE could now be placed in the correct envelope.
P(this letter is placed in an INCORRECT envelope) = 1/2. (Of the 2 remaining envelopes, only 1 is correct, so 1 is incorrect.)
P(last letter is placed in an INCORRECT envelope) = 1. (As noted above, only ONE of C and D could be placed in the correct envelope, so the other must be placed in an incorrect envelope.)

Since all of these events must happen in order for A to be the only envelope correctly placed, we multiply the fractions:
1/4 * 2/3 * 1/2 * 1.

Other cases:
Since there are 4 options for the one envelope correctly placed -- A, B, C, or D -- the result above must be multiplied by 4:
1/4 * 2/3 * 1/2 * 1 * 4 = 1/3.

Approach 3:

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):

Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Let the 4 letters be ABCD.

Total arrangements:
Total number of ways to arrange the 4 letters = 4! = 24.

Good arrangements:
In a good arrangement, EXACTLY ONE letter is in the correct position.
Number of options for the one letter put into the correct position = 4. (A, B, C, or D)
Number of ways to DERANGE the 3 remaining letters = 3! (1/2! - 1/3!) = 3-1 = 2.
To combine these options, we multiply:
4*2 = 8.

Good/total = 8/24 = 1/3.

[GMATinsight]

Tanya had 4 letters to send to 4 different addresses. For each letter she made an envelope with the correct address. If 4 letters are put in 4 envelopes randomly, what is the probability of only 1 letter in the correct envelope
1) 1/24
2) 1/8
3) 1/3
4) 3/8

OA is C
Thanks very much

Let's say the correct letter and envelop combination is
Letter --- Envelops
A --- P
B --- Q
C --- R
D --- S

UNDERSTANDING : Now we have to make sure that
If A goes in P, then
B Doesn't go in Q
C Doesn't go in R
D Doesn't go in S

Method:

Step 1: The Letter that Goes into Correct Envelop can be chosen in 4 ways (Either A or B or C or D)[Let's say A goes in Right Envelope

Step 2: In this case next Letter (Let's say B) has only 2 Choices (either R or S) (Let's say B goes in R)

Step 3: Third letter will have just 1 Choice (i.e. s only)

Step 4: Thourth letter will have just 1 Choice (i.e. Q only)

Total Ways to keep letters as asked = 4x2x1x1 = 8

Total ways to keep letters into envelopes = 4! = 24

Probability = 8/24 = 1/3

Answer: Option C

[Katy_]

Total probability: 4! = 24

No. of cases that there're 3 letters (or 4) put correctly: 4C3 x 1! = 4
No. of cases that there're 2 letters put correctly: 4C2 x 2! = 12

=> Answer = (24 - 4 - 12)/24 = 8/24 = 1/3

[Matt@VeritasPrep]

Total probability: 4! = 24

No. of cases that there're 3 letters (or 4) put correctly: 4C3 x 1! = 4
No. of cases that there're 2 letters put correctly: 4C2 x 2! = 12

=> Answer = (24 - 4 - 12)/24 = 8/24 = 1/3

An important note here: you can't have exactly three letters placed correctly. For instance, consider this arrangement of letters and envelopes, with three correct and the last one blank:

A B C D
A B C

There's only one place left to put letter D ... in envelope D!

So, interestingly enough, when you're placing these letters you can only place 0 right, 1 right, 2 right, or all 4 right. (This concept -- that if you've placed n items in n boxes, you can never have exactly (n - 1) placed correctly -- can come up on other questions, so it's worth mentioning.)