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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
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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.

Answer: D.
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Re: A certain company [#permalink]

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New post 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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Re: A certain company [#permalink]

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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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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
Answer is 8.

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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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New post 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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Re: Why D any not B? please help me out [#permalink]

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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


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
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Re: A certain company [#permalink]

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The best way to remember this is :
(Decisions) ^ (Players)
For this problem - 2 decisions , 3 players : 2^3=8

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Hi,
I am kind of lost on all probability type qs :/

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

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ashiima wrote:
Hi,
I am kind of lost on all probability type qs :/

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


Think in this way:
There is no restriction on the offices i.e. they can be vacant, they can accommodate all 3 employees etc. But there is a restriction on the employees i.e. each one of them must get an office.

Employee 1 can get an office in 2 ways - office A or office B
Employee 2 can get an office in 2 ways - office A or office B
Employee 3 can get an office in 2 ways - office A or office B
All three can be allotted offices in 2*2*2 = 8 ways
This takes care of all cases.
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The fastest way to solve this problem is by using the formula, 2^n, where n stands for the number of elements, or, in this case, the number of employees. This formula is derived from adding the number of combinations from

What’s important with this problem is not to treat it as a probability problem. While on the surface it may seem similar to a typical combinations problem, using the combinations formula to solve the problem is cumbersome.

Instead, use the formula, 2^n, where n stands for the number of elements, or, in this case, the number of employees.

This formula is derived from adding the number of combinations whenever you can select any number greater than zero and less than or equal to n. For instance, here we could have chosen any of three employees for the first office. So instead of using 3C0 + 3C1 + 3C2 + 3C3, we can use 2^3.

This formula becomes especially useful for larger numbers. Imagine the question were:

How many ways can 8 employees go in two offices?

(A) 8
(B) 32
(C) 48
(D) 64
(E) 120


Following the method of finding each case would take too much time. By using 2^n, we 2^8 = 64. (D)

Going back to my original point: do not think of this as a typical probability problem, but one that uses the 2^n concept. The problems, while not really a probability problem, employ the 2^n formula.

Positive integer N is the product of three distinct primes. How many factors in N?

Ans: 2^3 – 1 (zero is not a factor so hence we subtract one from 2^n).


A multiple-choice test has five possible answer choices. Any number of answers can be correct. (e.g. A-B-D is possible answer, C-D, or all five). How many different possible answers?

Ans: 2^5 – 1 = 31 You can’t leave question blank (like the empty office) so therefore -1.


By understanding the concept behind a question, instead of grouping a question under one general category, you should be able to solve problems more quickly.
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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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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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Re: A certain company assigns employees to offices in such a way [#permalink]

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

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New post 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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New post 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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