Problem Solving

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?


OE [spoiler]1/3[/spoiler]

Please explain.
Also what is derangement technique?
Is there any other way to solve the above question apart from the mentioned technique?

Let's call them letters A, B, C, and D, and envelopes A, B, C, and D.

Total ways 4 letters can be put into 4 envelopes.

4 x 3 x 2 x 1 = 24

Now to find the total ways one can be in the right place.

Start with letter A in envelope A, with the rest arranged in the other 3 envelopes such that none are in the right one.

There are two that could wrongly go into the first other envelope, and then for each of those two the other two have to both be in the wrong ones.

So envelope B could hold C or D. We get A C D B and A D B C. So the formula is 1 x 2 x 1 x 1 = 2

It works the same for B in B, 2.

Same for C in C, 2.

Same for D in D, 2.

So there are 8 ways to arrange with 1 in the right place.

So the chance of 1 being in the right place is [spoiler]8/24 = 1/3[/spoiler].

akash singhal wrote: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

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 = [spoiler]1/3[/spoiler].

The correct answer is D.

Alternate approach 1:

Let the 4 letters be A, B, C and D.

Case 1: Only A is placed in the correct envelope
P(A is placed in the correct envelope) = 1/4. (Of the 4 envelopes, only 1 corresponds to A.)
P(B is placed in a wrong envelope) = 2/3. (Of the 3 remaining envelopes, 2 are wrong.)
Of the 2 remaining letters, at most 1 can be now be placed in the correct envelope.
P(this letter is placed in the WRONG envelope) = 1/2. (Of the 2 remaining envelopes, 1 is wrong.)
P(last letter is placed in a wrong envelope) = 1/1 = 1. (The 1 remaining envelope must be wrong.)
To combine these probabilities, we multiply:
P(only A is placed in the correct envelope) = 1/4 * 2/3 * 1/2 * 1 = 1/12.

Using the same reasoning:
P(only B is placed in the correct envelope) = 1/12.
P(only C is placed in the correct envelope) = 1/12.
P(only D is placed in the correct envelope) = 1/12.

Since any of the probabilities above will yield a favorable outcome, we ADD the fractions:
1/12 + 1/12 + 1/12 + 1/12 = 4/12 = 1/3.

The correct answer is D.

Alternate approach 2:

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.

GMATGuruNY wrote:

akash singhal wrote: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

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 = [spoiler]1/3[/spoiler].

The correct answer is D.

Alternate approach 1:

Let the 4 letters be A, B, C and D.

Case 1: Only A is placed in the correct envelope
P(A is placed in the correct envelope) = 1/4. (Of the 4 envelopes, only 1 corresponds to A.)
P(B is placed in a wrong envelope) = 2/3. (Of the 3 remaining envelopes, 2 are wrong.)
Of the 2 remaining letters, at most 1 can be now be placed in the correct envelope.
P(this letter is placed in the WRONG envelope) = 1/2. (Of the 2 remaining envelopes, 1 is wrong.)
P(last letter is placed in a wrong envelope) = 1/1 = 1. (The 1 remaining envelope must be wrong.)
To combine these probabilities, we multiply:
P(only A is placed in the correct envelope) = 1/4 * 2/3 * 1/2 * 1 = 1/12.

Using the same reasoning:
P(only B is placed in the correct envelope) = 1/12.
P(only C is placed in the correct envelope) = 1/12.
P(only D is placed in the correct envelope) = 1/12.

Since any of the probabilities above will yield a favorable outcome, we ADD the fractions:
1/12 + 1/12 + 1/12 + 1/12 = 4/12 = 1/3.

The correct answer is D.

Alternate approach 2:

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.

Thanks Mitch for the alternative approaches.
Your insights are always helpful.

I think this one makes the most sense visually: here's a graph I drew up.

Yikes, that didn't work out well: here's a link if the image got cut off in your browser.

akash singhal wrote:Also what is derangement technique?
Is there any other way to solve the above question apart from the mentioned technique?

It's cool that there are and and it's cool to see so many ways to solve this. At the same time, the GMAT is not a math exam, or, beyond that, some kind of combinatorics test.

So the general answer to your question is that if a question is an official GMAT question or an accurate representation of an official GMAT question, then answering it will not require knowing some arcane technique.

While there are some things that one needs to review or learn in order to score high on the GMAT, the general idea of the test is not to get people to learn a huge amount of relatively useless stuff that they will never again use once they are done with the test. So answering even the most challenging GMAT combinatorics questions, for instance, requires knowing merely basic combinatorics concepts and approaches, which can be applied in creative ways to finding the answers.