Problem Solving

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Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

# Total arrangements to put 4 letters in 4 envelopes = 4! = 24
Ways to put only 1 letter in correct envelope = 8
ACDB --- ADBC --- CBDA --- DBAC --- BDCA --- DACB --- CABD and BCAD
Reqd prob = 8/24 = 1/3

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

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.

The correct answer is D.

The following solution does not require knowledge of the derangement formula:

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.

Another approach

A B C D are latters
a b c d are envlops
Total no of possiblities are 4!
for A fix a because we can put 1 letter in right envelope
now for B we have two envlops c or d lets fix d for it
now for C we have only one option left which is b
and for D we have only one envelope left c

Required Probility = favourable/ Total = 1x2x1x1/24 * 4 (We could put either A,BC or D in right envelope)= 1/3