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A certain company assigns employees to offices in such a way [#permalink]
08 Jan 2010, 13:18
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Difficulty:
65% (hard)
Question Stats:
50% (02:01) correct
50% (01:18) wrong based on 182 sessions
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?
Re: A certain company [#permalink]
08 Jan 2010, 15:58
2
This post received KUDOS
Expert's post
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?
Re: A certain company [#permalink]
09 Jan 2010, 03:25
2
This post received KUDOS
Expert's post
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.
Re: A certain company employee [#permalink]
25 Oct 2010, 03:38
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
Re: Why D any not B? please help me out [#permalink]
17 Nov 2010, 07:56
Expert's post
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
thanks in advance!!!
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 _________________
Re: A certain company assigns employees to offices in such a way [#permalink]
06 Jul 2013, 16:59
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! _________________
Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James
Re: A certain company assigns employees to offices in such a way [#permalink]
07 Jul 2013, 00:13
1
This post received KUDOS
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1
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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}). _________________
Re: A certain company assigns employees to offices in such a way [#permalink]
07 Jul 2013, 12:35
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 \(3^5\) _________________
Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James
Re: A certain company assigns employees to offices in such a way [#permalink]
10 Jul 2013, 11:40
Applying combinations I think would be in this way: \(C^4_1 * C^2_1 = 4*2 = 8\) _________________
Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James
Re: A certain company assigns employees to offices in such a way [#permalink]
31 Aug 2013, 06:47
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? _________________
Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!
Re: A certain company assigns employees to offices in such a way [#permalink]
03 Sep 2013, 05:42
Expert's post
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. _________________
Re: A certain company assigns employees to offices in such a way [#permalink]
15 Sep 2014, 18:24
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