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In how many ways can 6 different letters be posted in 3 letterboxes su

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In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   04 Jun 2020, 23:26
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Question Stats:

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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

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D
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In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   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



GMATinsight wrote:
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



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Re: In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   05 Jun 2020, 19:34
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First solution
4,1,1
Can be taken in 3 ways
Now letters can be chosen in
6C4×2C1×1C1= 15×2×1=30
Total number of ways =30×3=90
Second solution
1,2,3
Can be taken in 6 ways
Now letters can be chosen in
6C1×5C2×3C3= 6×10×1= 60
Total number of ways =6×60=360
Third solution
2,2,2
Can be taken in 1 way
Now letters can be chosen in
6C2×4C2×2C2= 15×6×1 =90
Total number of ways =90×1 =90
Total of all possible solutions=
90+360+90=540
Option D is the answer

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In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   05 Jun 2020, 22:47
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GMATinsight wrote:
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

Source: https://www.GMATinsight.com


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
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Re: In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   05 Jun 2020, 23:20
GMATinsight wrote:
GMATinsight wrote:
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



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


Can we not use partition concept of combination in this case ..
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Re: In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   05 Jun 2020, 23:57
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DasAshishAshutosh

Partition rule for the similar objects to be arranged.

this question specifically mentions that 6 letters are different so NO, we can not use partition method here. :)

Hope it help! :)

DasAshishAshutosh wrote:
GMATinsight wrote:
GMATinsight wrote:
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



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


Can we not use partition concept of combination in this case ..


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Re: In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   06 Jun 2020, 00:34
GMATinsight wrote:
DasAshishAshutosh

Partition rule for the similar objects to be arranged.

this question specifically mentions that 6 letters are different so NO, we can not use partition method here. :)

Hope it help! :)

Yes . Thanks GMATinsight
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Re: In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   13 Jul 2020, 22:41
GMATinsight wrote:
GMATinsight wrote:
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



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


Hi GMATinsight
Please explain Method 2, where we are subtracting 1 from 2^6. I cannot see how that case is already included in 3C1

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Re: In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink] New post   14 Jul 2020, 12:36
Expert Reply
Gokul20 wrote:
GMATinsight wrote:
GMATinsight wrote:
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



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


Hi GMATinsight
Please explain Method 2, where we are subtracting 1 from 2^6. I cannot see how that case is already included in 3C1

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Gokul20

Unwanted case in method 2 is "Exactly one box gets 0 letter"

3C1 = ways of choosing the box which gets 0 letter

while we calculate 2^6, it also includes a case in which all letters have gone in exactly one of the remaining two boxes i.e. one more box gets 0 letters. we need to exclude that case therefore we subtract that 1 case.
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Re: In how many ways can 6 different letters be posted in 3 letterboxes su [#permalink]
14 Jul 2020, 12:36
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