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  # In how many ways can 5 different toys be packed in 3
  
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  In how many ways can 5 different toys be packed in 3 [#permalink]  09 Nov 2006, 03:36
  00:00
  
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  Question Stats:
  
  0% (00:00) correct 0% (00:00) wrong based on 0 sessions
  In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
  
  a.20
  b.30
  c.50
  d.600
  e.1000
  Senior Manager
  Joined: 01 Sep 2006
  Posts: 302
  Location: Phoenix, AZ, USA
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  Box A B C
  1 1 3 can be 3 ways= 15
  1 2 2 can be 3 ways = 15
  
  Total 30
  Senior Manager
  Joined: 05 Oct 2006
  Posts: 268
  Followers: 1
  
  Kudos [?]: 4 [0], given: 0
  
  Re: PS Permutation [#permalink]  09 Nov 2006, 10:23
  each of the toy can goto any of the 3 boxes a,b or c...in 3 ways..
  so total no of ways in which 5 toys can be assigned to 3 boxes is
  
  3*3*3..5 times = 243...now this includes cases these toys might not goto one or 2 boxes..
  so i guess the ans should be not d or e..!!!
  i wud go with choice c.
  
  gmat strategy......hee hee hee...
  
  somebody show the way plz.
  SVP
  Joined: 01 May 2006
  Posts: 1805
  Followers: 8
  
  Kudos [?]: 99 [0], given: 0
  
  (C) for me
  
  The 3 boxes are a kind of illusion. It's just a way to combine toys together.
  
  So we can do 2 kinds of set:
  > 2 2 1 (we separate in 3 groups : 2 toys, 2 toys and 1 toy)
  > 3 1 1
  
  (2C5)*(2C3) + (3C5)*(1C2)
  = 5!/(3!*2!)*3 + 5!/(3!*2!)*2
  = 5*2*3 + 5*2*2
  = 50
  
  Last edited by Fig on 09 Nov 2006, 23:16, edited 1 time in total.
  Director
  Joined: 02 Mar 2006
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  Location: France
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  Thanks fig, OA is C.
  
  But can you please explain the second situation 3 1 1
  
  That means that we calculate the number of ways to take 3 toys from 5 and put them in the box. So 2 toys and 2 empty identical boxes are remaining.
  That's what disturbs me : you say there are two ways to put these 2 toys in the two boxes, or 1C2. But if the boxes are the same, let say A and A', it doesn't matter that one of the toy (X) goes in A or A'. I better explain with following:
  
  X is in A and Y is in A' OR X is in A' and Y is in A. since A=A', there is finally only one possibility:
  
  Like : How many words can I write with the letters A and A: only one! and not two(AA' or A'A).
  
  I hope you will understand my problem!
  SVP
  Joined: 01 May 2006
  Posts: 1805
  Followers: 8
  
  Kudos [?]: 99 [0], given: 0
  
  karlfurt wrote:
  Thanks fig, OA is C.
  
  But can you please explain the second situation 3 1 1
  
  That means that we calculate the number of ways to take 3 toys from 5 and put them in the box. So 2 toys and 2 empty identical boxes are remaining.
  That's what disturbs me : you say there are two ways to put these 2 toys in the two boxes, or 1C2. But if the boxes are the same, let say A and A', it doesn't matter that one of the toy (X) goes in A or A'. I better explain with following:
  
  X is in A and Y is in A' OR X is in A' and Y is in A. since A=A', there is finally only one possibility:
  
  Like : How many words can I write with the letters A and A: only one! and not two(AA' or A'A).
  
  I hope you will understand my problem!
  
  Well, I think the bold is not correct... The fact that A=A' in their propertities but does not mean we have only 1 possibility ; A and A' are 2 toys .... We need to select one to put it in the second box
  
  It's where boxes are important (and no more an illusion ) .... We have to select 1 toy among 2 remaing, identical toys (combination) to put in the second box
  
  Last edited by Fig on 09 Nov 2006, 23:16, edited 1 time in total.
  Intern
  Joined: 08 Nov 2006
  Posts: 36
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  I am on a different track here. The boxes are distinct, so you can't just have 221 and 311, you would also need 212, 113, etc. If you were just figuring out the different ways you could arrange 5 toys into 3 boxes, you would have 60 different ways. In each of those ways you have 2 toys left over to spread around the boxes, so I think you need to multiply 60 X the number of ways you can distribute 2 toys between the 3 boxes. Which is giving me 60X9 and not one of the answers, so I am missing something. But if it was the gmat, I would just take the answer 600 and move on .
  Director
  Joined: 02 Mar 2006
  Posts: 582
  Location: France
  Followers: 1
  
  Kudos [?]: 35 [0], given: 0
  
  Thanks! While writing a reply because it was still not clear, I understood!
  
  PS : I think you meant toys and not tools.
  
  Cheers
  VP
  Joined: 25 Jun 2006
  Posts: 1175
  Followers: 2
  
  Kudos [?]: 72 [0], given: 0
  
  andrewnorway wrote:
  I am on a different track here. The boxes are distinct, so you can't just have 221 and 311, you would also need 212, 113, etc. If you were just figuring out the different ways you could arrange 5 toys into 3 boxes, you would have 60 different ways. In each of those ways you have 2 toys left over to spread around the boxes, so I think you need to multiply 60 X the number of ways you can distribute 2 toys between the 3 boxes. Which is giving me 60X9 and not one of the answers, so I am missing something. But if it was the gmat, I would just take the answer 600 and move on .
  
  identical boxes!.
  
  it is more about classification as Fig explained.
  SVP
  Joined: 01 May 2006
  Posts: 1805
  Followers: 8
  
  Kudos [?]: 99 [0], given: 0
  
  karlfurt wrote:
  Thanks! While writing a reply because it was still not clear, I understood!
  
  PS : I think you meant toys and not tools.
  
  Cheers
  
  Ahhh... Late readings :p... Absolutely... They are toys not tools
  
  Sorry on this confusing mistake.... I change all my posts now .
  Similar topics Replies Last post
  Similar
  Topics:
  16 In how many ways can 5 different marbles be distributed in 4 10 27 Apr 2014, 23:59
  9 How many ways can 5 different colored marbles be placed in 3 2 14 Oct 2013, 07:53
  2 In how many ways 5 different balls can be arranged in to 3 6 07 Aug 2010, 05:10
  In how many ways can 3 cards be selected from a pack of 52 1 26 Nov 2005, 12:17
  In how many ways can 4 identical red balls and 5 different 10 05 Oct 2005, 14:47
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distilabel_metadata.content: Find all School-related info fast with the new School-Specific MBA Forum
  
   It is currently 04 May 2015, 16:53
  
  ### GMAT Club Daily Prep
  
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  # Events & Promotions
  
  ###### Events & Promotions in June
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  # In how many ways can 5 different toys be packed in 3
  
  Author Message
  TAGS:
  Director
  Joined: 02 Mar 2006
  Posts: 582
  Location: France
  Followers: 1
  
  Kudos [?]: 35 [0], given: 0
  
  In how many ways can 5 different toys be packed in 3 [#permalink]  09 Nov 2006, 03:36
  00:00
  
  Difficulty:
  
  (N/A)
  
  Question Stats:
  
  0% (00:00) correct 0% (00:00) wrong based on 0 sessions
  In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
  
  a.20
  b.30
  c.50
  d.600
  e.1000
  Senior Manager
  Joined: 01 Sep 2006
  Posts: 302
  Location: Phoenix, AZ, USA
  Followers: 1
  
  Kudos [?]: 15 [0], given: 0
  
  Box A B C
  1 1 3 can be 3 ways= 15
  1 2 2 can be 3 ways = 15
  
  Total 30
  Senior Manager
  Joined: 05 Oct 2006
  Posts: 268
  Followers: 1
  
  Kudos [?]: 4 [0], given: 0
  
  Re: PS Permutation [#permalink]  09 Nov 2006, 10:23
  each of the toy can goto any of the 3 boxes a,b or c...in 3 ways..
  so total no of ways in which 5 toys can be assigned to 3 boxes is
  
  3*3*3..5 times = 243...now this includes cases these toys might not goto one or 2 boxes..
  so i guess the ans should be not d or e..!!!
  i wud go with choice c.
  
  gmat strategy......hee hee hee...
  
  somebody show the way plz.
  SVP
  Joined: 01 May 2006
  Posts: 1805
  Followers: 8
  
  Kudos [?]: 99 [0], given: 0
  
  (C) for me
  
  The 3 boxes are a kind of illusion. It's just a way to combine toys together.
  
  So we can do 2 kinds of set:
  > 2 2 1 (we separate in 3 groups : 2 toys, 2 toys and 1 toy)
  > 3 1 1
  
  (2C5)*(2C3) + (3C5)*(1C2)
  = 5!/(3!*2!)*3 + 5!/(3!*2!)*2
  = 5*2*3 + 5*2*2
  = 50
  
  Last edited by Fig on 09 Nov 2006, 23:16, edited 1 time in total.
  Director
  Joined: 02 Mar 2006
  Posts: 582
  Location: France
  Followers: 1
  
  Kudos [?]: 35 [0], given: 0
  
  Thanks fig, OA is C.
  
  But can you please explain the second situation 3 1 1
  
  That means that we calculate the number of ways to take 3 toys from 5 and put them in the box. So 2 toys and 2 empty identical boxes are remaining.
  That's what disturbs me : you say there are two ways to put these 2 toys in the two boxes, or 1C2. But if the boxes are the same, let say A and A', it doesn't matter that one of the toy (X) goes in A or A'. I better explain with following:
  
  X is in A and Y is in A' OR X is in A' and Y is in A. since A=A', there is finally only one possibility:
  
  Like : How many words can I write with the letters A and A: only one! and not two(AA' or A'A).
  
  I hope you will understand my problem!
  SVP
  Joined: 01 May 2006
  Posts: 1805
  Followers: 8
  
  Kudos [?]: 99 [0], given: 0
  
  karlfurt wrote:
  Thanks fig, OA is C.
  
  But can you please explain the second situation 3 1 1
  
  That means that we calculate the number of ways to take 3 toys from 5 and put them in the box. So 2 toys and 2 empty identical boxes are remaining.
  That's what disturbs me : you say there are two ways to put these 2 toys in the two boxes, or 1C2. But if the boxes are the same, let say A and A', it doesn't matter that one of the toy (X) goes in A or A'. I better explain with following:
  
  X is in A and Y is in A' OR X is in A' and Y is in A. since A=A', there is finally only one possibility:
  
  Like : How many words can I write with the letters A and A: only one! and not two(AA' or A'A).
  
  I hope you will understand my problem!
  
  Well, I think the bold is not correct... The fact that A=A' in their propertities but does not mean we have only 1 possibility ; A and A' are 2 toys .... We need to select one to put it in the second box
  
  It's where boxes are important (and no more an illusion ) .... We have to select 1 toy among 2 remaing, identical toys (combination) to put in the second box
  
  Last edited by Fig on 09 Nov 2006, 23:16, edited 1 time in total.
  Intern
  Joined: 08 Nov 2006
  Posts: 36
  Followers: 0
  
  Kudos [?]: 1 [0], given: 0
  
  I am on a different track here. The boxes are distinct, so you can't just have 221 and 311, you would also need 212, 113, etc. If you were just figuring out the different ways you could arrange 5 toys into 3 boxes, you would have 60 different ways. In each of those ways you have 2 toys left over to spread around the boxes, so I think you need to multiply 60 X the number of ways you can distribute 2 toys between the 3 boxes. Which is giving me 60X9 and not one of the answers, so I am missing something. But if it was the gmat, I would just take the answer 600 and move on .
  Director
  Joined: 02 Mar 2006
  Posts: 582
  Location: France
  Followers: 1
  
  Kudos [?]: 35 [0], given: 0
  
  Thanks! While writing a reply because it was still not clear, I understood!
  
  PS : I think you meant toys and not tools.
  
  Cheers
  VP
  Joined: 25 Jun 2006
  Posts: 1175
  Followers: 2
  
  Kudos [?]: 72 [0], given: 0
  
  andrewnorway wrote:
  I am on a different track here. The boxes are distinct, so you can't just have 221 and 311, you would also need 212, 113, etc. If you were just figuring out the different ways you could arrange 5 toys into 3 boxes, you would have 60 different ways. In each of those ways you have 2 toys left over to spread around the boxes, so I think you need to multiply 60 X the number of ways you can distribute 2 toys between the 3 boxes. Which is giving me 60X9 and not one of the answers, so I am missing something. But if it was the gmat, I would just take the answer 600 and move on .
  
  identical boxes!.
  
  it is more about classification as Fig explained.
  SVP
  Joined: 01 May 2006
  Posts: 1805
  Followers: 8
  
  Kudos [?]: 99 [0], given: 0
  
  karlfurt wrote:
  Thanks! While writing a reply because it was still not clear, I understood!
  
  PS : I think you meant toys and not tools.
  
  Cheers
  
  Ahhh... Late readings :p... Absolutely... They are toys not tools
  
  Sorry on this confusing mistake.... I change all my posts now .
  Similar topics Replies Last post
  Similar
  Topics:
  16 In how many ways can 5 different marbles be distributed in 4 10 27 Apr 2014, 23:59
  9 How many ways can 5 different colored marbles be placed in 3 2 14 Oct 2013, 07:53
  2 In how many ways 5 different balls can be arranged in to 3 6 07 Aug 2010, 05:10
  In how many ways can 3 cards be selected from a pack of 52 1 26 Nov 2005, 12:17
  In how many ways can 4 identical red balls and 5 different 10 05 Oct 2005, 14:47
  Display posts from previous: Sort by
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