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A certain company assigns employees to offices in such a way [#permalink]

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08 Jan 2010, 14:18

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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 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?

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.

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

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]

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06 Jul 2013, 17:59

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!
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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!

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]

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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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Re: A certain company assigns employees to offices in such a way [#permalink]

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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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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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Re: A certain company assigns employees to offices in such a way [#permalink]

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15 Sep 2014, 19:24

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Re: A certain company assigns employees to offices in such a way [#permalink]

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18 Apr 2016, 12:00

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A certain company assigns employees to offices in such a way [#permalink]

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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

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