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  A certain company assigns employees to offices in such a way
  
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  A certain company assigns employees to offices in such a way [#permalink]
  
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  A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
  
  A. 5
  B. 6
  C. 7
  D. 8
  E. 9
  [Reveal] Spoiler: OA
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  08 Jan 2010, 16:58
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  sagarsabnis wrote:
  A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
  
  A. 5
  B. 6
  C. 7
  D. 8
  E. 9
  [Reveal] Spoiler:
  D
  
  Each of three employee can be assigned to either of offices, meaning that each has 2 choices --> 2*2*2=2^3=8.
  
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  09 Jan 2010, 03:27
  i am still not able to understand. Can you please explain in detail?
  
  also please tell me where i went wrong.This was my logic.
  
  No. of people
  office 1: 0|0|0|1|1|1|2|2|3
  office 2: 1|2|3|0|1|2|0|1|0
  
  this gives me 9 possible combination
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  09 Jan 2010, 04:25
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  sagarsabnis wrote:
  i am still not able to understand. Can you please explain in detail?
  
  also please tell me where i went wrong.This was my logic.
  
  No. of people
  office 1: 0|0|0|1|1|1|2|2|3
  office 2: 1|2|3|0|1|2|0|1|0
  
  this gives me 9 possible combination
  
  First of all you should assign ALL 3 employees to either of the offices. You can have the following scenarios:
  
  No. of people
  ***********A|B|C|D|
  office 1: 0|1|2|3|
  office 2: 3|2|1|0|
  
  In scenario (A) and (D) there is only one way to assign three people. But in (B) and (C) there will be 3 cases in each:
  
  Let's say there are 3 employees: Tom, Mary and Kate. In (B): Tom can be in office #1 and Mary/Kate in #2 OR Mary can be in #1 and Tom/Kate in #2 OR Kate in #1 and Tom/Mary in #2. Total 3 cases for (B). The same for (C). (A)+(B)+(C)+(D)=1+3+3+1=8.
  
  The way I solved this was different:
  
  Each of the three employees, Tom, Mary and Kate, has two choices office #1 or office #2. Hence total # of combinations (assignments) is 2*2*2=2^3=8.
  
  Hope it's clear.
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  25 Oct 2010, 04:26
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  Two offices can be filled in two ways, when all the three employee will be in same room or when two employee in one room and one in other room.
  
  when all the three employee will be in same = 3C3 * 2! =2 (2!, because any of the room can be taken)
  
  when two employee in one room and one in other room. = 3C2 * 1C1 * 2! = 6
  
  Hence total ways = 6+2
  
  Consider KUDOS if it is helpful to u.
  
  Last edited by ankitranjan on 25 Oct 2010, 04:43, edited 1 time in total.
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  25 Oct 2010, 04:38
  monirjewel wrote:
  A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office, In how many ways can the company assign 3 employees to 2 different offices?
  
  A) 5
  B) 6
  C) 7
  D) 8
  E) 9
  
  Every employee has got the possibilit of getting assigned to any of the two offices.
  Hence total possibilities = 2^3 = 8
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  17 Nov 2010, 08:56
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  SoniaSaini wrote:
  A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
  A. 5
  B. 6
  C. 7
  D. 8
  E. 9
  
  For each one of the 3 employees, there are two choices. He can be allotted to any one of the two offices. Hence total number ways will be 2 * 2* 2 = 8 ways
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  21 Nov 2010, 07:12
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  The best way to remember this is :
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  For this problem - 2 decisions , 3 players : 2^3=8
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  Re: A certain company assigns employees to offices in such a way [#permalink]
  
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  06 Jul 2013, 17:59
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  Thank you. Then, if the company assigns employees to offices in such a way that if the offices can not be empty and more than one employee can be assigned to an office. And we have 5 employees and 3 rooms, the answer would be:
  
  120?
  I mean, 5!
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  07 Jul 2013, 01:13
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  Maxirosario2012 wrote:
  Thank you. Then, if the company assigns employees to offices in such a way that if the offices can not be empty and more than one employee can be assigned to an office. And we have 5 employees and 3 rooms, the answer would be:
  
  120?
  I mean, 5!
  
  No. It would be 3^5 minus restriction.
  
  For example, for 5 employees and 2 offices it would be 2^5 - 2 ({5-0} and {0-5}).
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  07 Jul 2013, 13:35
  Thank you Bunuel!
  I have difficulties learning combinations, this is my weakest area in the GMAT. I am planning to practice all the combinations problems in the forum.
  Regarding the problem that I have posted before, I think that you mean:
  
  $$3^5$$ - the combinations in which zero is an element in the set and it cannot be zero in any of the slots, with the restrictions that the 3 elements must sum up 5):
  {(005),(014),(023),(032),(041) ; (050)(140),(230),(320),(410) ; (500),(104),(203),(302),(401)}
  
  243 - 15 = 228
  
  I tried to apply combinatorics formulas to this problem (because writing that set is very time consuming) but I could not figure it out.
  Translating the problem:
  I need to find the number of combinations of three digits in which at least one of the digits is "0", the sum of those three digits is 5 and the digits range from 0 to 5 (six elements).
  Then, substract this number from $$3^5$$
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  10 Jul 2013, 12:40
  Applying combinations I think would be in this way:
  $$C^4_1 * C^2_1 = 4*2 = 8$$
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  31 Aug 2013, 07:47
  ok, can someone tell me what's wrong with my thinking..
  1st office can have any 3 employees.. therefore 3 options,
  2nd office can also have any of 3 employees hence again 3 options
  so it should be 3*3=9
  
  i think the logic is similar to the way Bunuel did..the only difference is in that case we had 2 choices for each employee therefore it was 2*2*2=8.. but why is the answer different in both cases?
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  03 Sep 2013, 06:42
  nikhil007 wrote:
  ok, can someone tell me what's wrong with my thinking..
  1st office can have any 3 employees.. therefore 3 options,
  2nd office can also have any of 3 employees hence again 3 options
  so it should be 3*3=9
  
  i think the logic is similar to the way Bunuel did..the only difference is in that case we had 2 choices for each employee therefore it was 2*2*2=8.. but why is the answer different in both cases?
  
  We are distributing employees to the offices not vise-versa.
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  18 Apr 2016, 12:28
  Three people that can go in either office 1 or 2
  So 2^3 =8
  
  Important is to see that you are not distributing offices to people. That would be 3^2 instead of 2^3
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  18 Apr 2016, 15:46
  with 2 offices and 3 employees,
  there are 3 ways to have one's own office,
  3 ways to share an office with one other, and
  2 ways to share an office with two others, or
  8 ways total
  Attachments
  
  office.jpg [ 34.22 KiB | Viewed 2228 times ]
  
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  09 May 2016, 04:07
  heh got that question in CAT 5 of EP2
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  Re: A certain company assigns employees to offices in such a way   [#permalink] 09 May 2016, 04:07
  
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explanation: The document you provided contains a math question related to the assignment of employees to offices. The text describes a scenario where a company assigns 3 employees to 2 different offices, with the possibility of offices being empty and more than one employee being assigned to an office. The question asks for the number of ways the company can assign the employees to the offices.
  
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  A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
  
  A. 5
  B. 6
  C. 7
  D. 8
  E. 9
  [Reveal] Spoiler: OA
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  08 Jan 2010, 16:58
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  sagarsabnis wrote:
  A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
  
  A. 5
  B. 6
  C. 7
  D. 8
  E. 9
  [Reveal] Spoiler:
  D
  
  Each of three employee can be assigned to either of offices, meaning that each has 2 choices --> 2*2*2=2^3=8.
  
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  09 Jan 2010, 03:27
  i am still not able to understand. Can you please explain in detail?
  
  also please tell me where i went wrong.This was my logic.
  
  No. of people
  office 1: 0|0|0|1|1|1|2|2|3
  office 2: 1|2|3|0|1|2|0|1|0
  
  this gives me 9 possible combination
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  09 Jan 2010, 04:25
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  sagarsabnis wrote:
  i am still not able to understand. Can you please explain in detail?
  
  also please tell me where i went wrong.This was my logic.
  
  No. of people
  office 1: 0|0|0|1|1|1|2|2|3
  office 2: 1|2|3|0|1|2|0|1|0
  
  this gives me 9 possible combination
  
  First of all you should assign ALL 3 employees to either of the offices. You can have the following scenarios:
  
  No. of people
  ***********A|B|C|D|
  office 1: 0|1|2|3|
  office 2: 3|2|1|0|
  
  In scenario (A) and (D) there is only one way to assign three people. But in (B) and (C) there will be 3 cases in each:
  
  Let's say there are 3 employees: Tom, Mary and Kate. In (B): Tom can be in office #1 and Mary/Kate in #2 OR Mary can be in #1 and Tom/Kate in #2 OR Kate in #1 and Tom/Mary in #2. Total 3 cases for (B). The same for (C). (A)+(B)+(C)+(D)=1+3+3+1=8.
  
  The way I solved this was different:
  
  Each of the three employees, Tom, Mary and Kate, has two choices office #1 or office #2. Hence total # of combinations (assignments) is 2*2*2=2^3=8.
  
  Hope it's clear.
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  Re: A certain company employee [#permalink]
  
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  25 Oct 2010, 04:26
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  Two offices can be filled in two ways, when all the three employee will be in same room or when two employee in one room and one in other room.
  
  when all the three employee will be in same = 3C3 * 2! =2 (2!, because any of the room can be taken)
  
  when two employee in one room and one in other room. = 3C2 * 1C1 * 2! = 6
  
  Hence total ways = 6+2
  
  Consider KUDOS if it is helpful to u.
  
  Last edited by ankitranjan on 25 Oct 2010, 04:43, edited 1 time in total.
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  Re: A certain company employee [#permalink]
  
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  25 Oct 2010, 04:38
  monirjewel wrote:
  A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office, In how many ways can the company assign 3 employees to 2 different offices?
  
  A) 5
  B) 6
  C) 7
  D) 8
  E) 9
  
  Every employee has got the possibilit of getting assigned to any of the two offices.
  Hence total possibilities = 2^3 = 8
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  17 Nov 2010, 08:56
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  SoniaSaini wrote:
  A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
  A. 5
  B. 6
  C. 7
  D. 8
  E. 9
  
  For each one of the 3 employees, there are two choices. He can be allotted to any one of the two offices. Hence total number ways will be 2 * 2* 2 = 8 ways
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  21 Nov 2010, 07:12
  2
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  The best way to remember this is :
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  For this problem - 2 decisions , 3 players : 2^3=8
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  01 Oct 2012, 13:13
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  Re: A certain company assigns employees to offices in such a way [#permalink]
  
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  06 Jul 2013, 17:59
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  Thank you. Then, if the company assigns employees to offices in such a way that if the offices can not be empty and more than one employee can be assigned to an office. And we have 5 employees and 3 rooms, the answer would be:
  
  120?
  I mean, 5!
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  07 Jul 2013, 01:13
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  Maxirosario2012 wrote:
  Thank you. Then, if the company assigns employees to offices in such a way that if the offices can not be empty and more than one employee can be assigned to an office. And we have 5 employees and 3 rooms, the answer would be:
  
  120?
  I mean, 5!
  
  No. It would be 3^5 minus restriction.
  
  For example, for 5 employees and 2 offices it would be 2^5 - 2 ({5-0} and {0-5}).
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  07 Jul 2013, 13:35
  Thank you Bunuel!
  I have difficulties learning combinations, this is my weakest area in the GMAT. I am planning to practice all the combinations problems in the forum.
  Regarding the problem that I have posted before, I think that you mean:
  
  $$3^5$$ - the combinations in which zero is an element in the set and it cannot be zero in any of the slots, with the restrictions that the 3 elements must sum up 5):
  {(005),(014),(023),(032),(041) ; (050)(140),(230),(320),(410) ; (500),(104),(203),(302),(401)}
  
  243 - 15 = 228
  
  I tried to apply combinatorics formulas to this problem (because writing that set is very time consuming) but I could not figure it out.
  Translating the problem:
  I need to find the number of combinations of three digits in which at least one of the digits is "0", the sum of those three digits is 5 and the digits range from 0 to 5 (six elements).
  Then, substract this number from $$3^5$$
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  Re: A certain company assigns employees to offices in such a way [#permalink]
  
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  10 Jul 2013, 12:40
  Applying combinations I think would be in this way:
  $$C^4_1 * C^2_1 = 4*2 = 8$$
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  31 Aug 2013, 07:47
  ok, can someone tell me what's wrong with my thinking..
  1st office can have any 3 employees.. therefore 3 options,
  2nd office can also have any of 3 employees hence again 3 options
  so it should be 3*3=9
  
  i think the logic is similar to the way Bunuel did..the only difference is in that case we had 2 choices for each employee therefore it was 2*2*2=8.. but why is the answer different in both cases?
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  03 Sep 2013, 06:42
  nikhil007 wrote:
  ok, can someone tell me what's wrong with my thinking..
  1st office can have any 3 employees.. therefore 3 options,
  2nd office can also have any of 3 employees hence again 3 options
  so it should be 3*3=9
  
  i think the logic is similar to the way Bunuel did..the only difference is in that case we had 2 choices for each employee therefore it was 2*2*2=8.. but why is the answer different in both cases?
  
  We are distributing employees to the offices not vise-versa.
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  15 Sep 2014, 19:24
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  18 Apr 2016, 12:00
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  18 Apr 2016, 12:28
  Three people that can go in either office 1 or 2
  So 2^3 =8
  
  Important is to see that you are not distributing offices to people. That would be 3^2 instead of 2^3
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  A certain company assigns employees to offices in such a way [#permalink]
  
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  18 Apr 2016, 15:46
  with 2 offices and 3 employees,
  there are 3 ways to have one's own office,
  3 ways to share an office with one other, and
  2 ways to share an office with two others, or
  8 ways total
  Attachments
  
  office.jpg [ 34.22 KiB | Viewed 2228 times ]
  
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  09 May 2016, 04:07
  heh got that question in CAT 5 of EP2
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  Re: A certain company assigns employees to offices in such a way   [#permalink] 09 May 2016, 04:07
  
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document you provided contains a math question related to the assignment of employees to offices. The text describes a scenario where a company assigns 3 employees to 2 different offices, with the possibility of offices being empty and more than one employee being assigned to an office. The question asks for the number of ways the company can assign the employees to the offices.
  
  1. The document contains a math question.
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  Label: question
  