Author 
Message 
TAGS:

Hide Tags

GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4350
Location: India
GMAT: QUANT EXPERT
WE: Education (Education)

In how many ways can 6 different letters be posted in 3 letterboxes su
[#permalink]
Show Tags
04 Jun 2020, 22:26
Question Stats:
52% (02:16) correct 48% (02:06) wrong based on 25 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6442
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

In how many ways can 6 different letters be posted in 3 letterboxes su
[#permalink]
Show Tags
05 Jun 2020, 12:10
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 GMATinsightsir 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 Posted from my mobile device



Senior Manager
Joined: 18 Dec 2017
Posts: 300

Re: In how many ways can 6 different letters be posted in 3 letterboxes su
[#permalink]
Show Tags
05 Jun 2020, 18:34
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
Posted from my mobile device



GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4350
Location: India
GMAT: QUANT EXPERT
WE: Education (Education)

In how many ways can 6 different letters be posted in 3 letterboxes su
[#permalink]
Show Tags
05 Jun 2020, 21:47
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 Method1: Distribution of letters can happen in three ways (411) (222) and (321) ways Case 1: 321 = 6C3*3C2*3! = 360 Case 2: 222 = 6C2*4C2 = 15*6 = 90 Case 3: 411 = 6C4*3! = 90 Total Ways = 360+90+90 = 540Answer: Option D Method2: Archit3110Favourable 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*(2^61) = 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 Method3: 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
_________________



Stern School Moderator
Joined: 26 May 2020
Posts: 273
Concentration: General Management, Technology
WE: Analyst (Computer Software)

Re: In how many ways can 6 different letters be posted in 3 letterboxes su
[#permalink]
Show Tags
05 Jun 2020, 22:20
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 Method1: Distribution of letters can happen in three ways (411) (222) and (321) ways Case 1: 321 = 6C3*3C2*3! = 360 Case 2: 222 = 6C2*4C2 = 15*6 = 90 Case 3: 411 = 6C4*3! = 90 Total Ways = 360+90+90 = 540Answer: Option D Method2: Archit3110Favourable 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*(2^61) = 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 Method3: 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 ..
_________________
Thank you. Regards, Ashish A Das.
The more realistic you are during your practice, the more confident you will be during the CAT.



GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4350
Location: India
GMAT: QUANT EXPERT
WE: Education (Education)

Re: In how many ways can 6 different letters be posted in 3 letterboxes su
[#permalink]
Show Tags
05 Jun 2020, 22:57
DasAshishAshutoshPartition 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 Method1: Distribution of letters can happen in three ways (411) (222) and (321) ways Case 1: 321 = 6C3*3C2*3! = 360 Case 2: 222 = 6C2*4C2 = 15*6 = 90 Case 3: 411 = 6C4*3! = 90 Total Ways = 360+90+90 = 540Answer: Option D Method2: Archit3110Favourable 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*(2^61) = 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 Method3: 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 .. Posted from my mobile device
_________________



Stern School Moderator
Joined: 26 May 2020
Posts: 273
Concentration: General Management, Technology
WE: Analyst (Computer Software)

Re: In how many ways can 6 different letters be posted in 3 letterboxes su
[#permalink]
Show Tags
05 Jun 2020, 23:34
GMATinsight wrote: DasAshishAshutoshPartition 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
_________________
Thank you. Regards, Ashish A Das.
The more realistic you are during your practice, the more confident you will be during the CAT.




Re: In how many ways can 6 different letters be posted in 3 letterboxes su
[#permalink]
05 Jun 2020, 23:34




