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04 Jun 2020, 23:26
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:21) correct
58%
(02:10)
wrong
based on 182
sessions
History
Date
Time
Result
Not Attempted Yet
In how many ways can 6 different letters be posted in 3 letterboxes such that each box receives atleast 1 letter?
(A) 120
(B) 180
(C) 360
(D) 540
(E) 720
(A) 120
(B) 180
(C) 360
(D) 540
(E) 720
Source: https://www.GMATinsight.com
05 Jun 2020, 22:47
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GMATinsight
Method-1:
Distribution of letters can happen in three ways (4-1-1) (2-2-2) and (3-2-1) ways
Case 1: 3-2-1 = 6C3*3C2*3! = 360
Case 2: 2-2-2 = 6C2*4C2 = 15*6 = 90
Case 3: 4-1-1 = 6C4*3! = 90
Total Ways = 360+90+90 = 540
Answer: Option D
Method-2:
Archit3110
Favourable Cases = Total cases - Unwanted cases
Total ways of distribution of 6 letters = 3*3*3*3*3*3 = 729
Unwanted cases = Atleast one box gets zero letter = 3C1*(2^6-1) = 3*63 = 189
3C1 = Ways to select the box that gets zero letter
2^6 = Total ways of distributing 6 letters into remaining 2 boxes with each letter having 2 choices Letter box 1 or box 2
-1 = Removing a case in which only one box receives all letters (counted twice due to 3C1)
i.e. Favourable cases = 729 - 189 = 540
Answer: Option D
Method-3:
6,0,0 = 3 ways
5,1,0 = 3C2*6C5*2! = 36 ways
3,3,0 = 3C2*6C3 = 60 ways
4,2,0 = 3C2*6C4*2! = 90 ways
So unfavorable cases = 3+36+60+90 = 189
Total ways of distribution of 6 letters = 3*3*3*3*3*3 = 729
i.e. Favourable cases = 729 - 189 = 540
Answer: Option D
General Discussion
Archit3110
05 Jun 2020, 13:10
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Total possible arrangement to arrange 6 letters in 3 letter boxes such that each box has 1 letter
4,1,1 ; 3,2,1 ; 2,2,2
Ways to arrange
4,1,1= 6!/4!*1*1*2! = 15
3,2,1= 6!/3!*2! = 60
2,2,2= 6!/2!*2!*2!*3!= 15
Total 15+15+60;90
And letter boxes can be arranged in 3! Ways so 90*3!= 540
Option D
GMATinsight
sir please provide solution considering
(total cases- unwanted cases)
I tried solving making unwanted cases but not getting answer..
6,0,0
5,1,0
3,3,0
4,2,0
Posted from my mobile device
4,1,1 ; 3,2,1 ; 2,2,2
Ways to arrange
4,1,1= 6!/4!*1*1*2! = 15
3,2,1= 6!/3!*2! = 60
2,2,2= 6!/2!*2!*2!*3!= 15
Total 15+15+60;90
And letter boxes can be arranged in 3! Ways so 90*3!= 540
Option D
GMATinsight
sir please provide solution considering
(total cases- unwanted cases)
I tried solving making unwanted cases but not getting answer..
6,0,0
5,1,0
3,3,0
4,2,0
GMATinsight
Posted from my mobile device
