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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! _________________

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

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 \(3^5\) _________________

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]

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10 Jul 2013, 12:40

Applying combinations I think would be in this way: \(C^4_1 * C^2_1 = 4*2 = 8\) _________________

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]

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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? _________________

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!

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]

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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

Hello from the GMAT Club BumpBot!

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 _________________

Took the Gmat and got a 520 after studying for 3 weeks with a fulltime job. Now taking it again, but with 6 weeks of prep time and a part time job. Studying every day is key, try to do at least 5 exercises a day.

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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