Five Letters Are To Be Placed Into Five Addressed Envelopes GMAT Problem Solving

Question: Five letters are to be placed into five addressed envelopes.If the letters are placed into the envelopes randomly, in how many ways can the letters be placed so that none of the letters is in its corresponding envelope?

1. 44
2. 30
3. 120
4. 80
5. 85

“Five letters are to be placed into five addressed envelopes.”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “The Official Guide for GMAT Reviews”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

It is given in the question that five letters are to be placed into five addressed envelopes. If the letters are placed into the envelopes randomly. It is asked to find out in how many ways can the letters be placed so that none of the letters is in its corresponding envelope.
This is a question from permutation and combination,
No of derangements of objects if there are n objects,
= n! (1 -1/1! + 1/2! - 1/3! + ........(-1)n 1/n!)
Here we are given 5 envelopes so here n = 5
5! (1 -1/1! + 1/2! - 1/3! +1/4! - 1/5!)
= 120 (1/2 -1/6 + 1/24 - 1/120)
= 120/ 120 (60 -20 +5-1)
= 44

Correct Answer: A

Approach Solution 2:
It is given in the question that five letters are to be placed into five addressed envelopes. If the letters are placed into the envelopes randomly. It is asked to find out in how many ways can the letters be placed so that none of the letters is in its corresponding envelope.
Suppose ABCDE is the right order
Number of ways to put only A in the right place
i)When B is at 3rd place - ACBED, ADBEC, AEBCD
ii)When B is at 4th place - ACEBD, ADEBC, AEDBC
iii) When B is at 5th place - ACEDB, ADECB, AEDCB
There are total 9 ways (Not 3)
Total number of permutation = 5! =120
1 letter is placed correctly in the correct envelope = 5C1*9 = 45
Total no. of ways in which none of the letters are placed correctly = 120 - (1+45+20+10+0) = 44 ways

Correct Answer: A

Approach Solution 3:

Number of derangement if there are total n objects =n!(1−11!+12!−13!+..........+(−1)n1n!)=n!(1−11!+12!−13!+..........+(−1)n1n!)
Number of ways in which none of the letters put in its corresponding envelope
=5!(1−11!+12!−13!+14!−15!)=5!(1−11!+12!−13!+14!−15!)
=120(12−16+124−1120)=120(12−16+124−1120)
=120120(60−20+5−1)=120120(60−20+5−1)

=44

Correct Answer: A

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