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by alltimeacheiver » Fri Feb 11, 2011 1:39 am
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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by GMATGuruNY » Fri Feb 11, 2011 2:45 am
alltimeacheiver 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
We need to count only the number of ways that the 3 employees can be assigned to the first office, since whoever isn't assigned to the first office must automatically be assigned to the second office.

Given n elements, the number of ways to choose 0 or more of the n elements = 2^n.

Thus, given 3 people, the number of ways to choose 0 or more of them for the first office = 2^3 = 8.

We also could simply write out all the options for the first office. Given 3 people ABC, the following combinations could be assigned to the first office:

ABC
AB
AC
BC
A
B
C
None

Same answer: 8 ways.

The correct answer is D.
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by albinogmat » Mon Nov 21, 2011 4:24 am
I read in one of the maths books that the formula is 2^n-1 for number of ways of selecting one or more thing in n given things.
I chose C with that formula. Please clarify.
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by GMATGuruNY » Mon Nov 21, 2011 4:31 am
albinogmat wrote:I read in one of the maths books that the formula is 2^n-1 for number of ways of selecting one or more thing in n given things.
I chose C with that formula. Please clarify.
Yes, the number of ways to choose ONE OR MORE elements from n choices = (2^n) - 1.

In the problem above, it is possible to assign NO secretaries to the first office.
The number of ways to choose 0 OR MORE elements from n choices = 2^n.
Thus, the number of ways to assign 0 OR MORE of the 3 secretaries to the first office = 2³ = 8.
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by user123321 » Mon Nov 21, 2011 4:58 am
alltimeacheiver wrote: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
I am not so sure whether question is well formed. The answer will be 8 if they say any number of employees can be assigned to all the offices. Then only 2^3 should be answer. or am i missing something :(

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by tuanquang269 » Tue Nov 22, 2011 8:52 pm
Tricky in the words translation. I still do not understand the formula here:
Given n elements, the number of ways to choose 0 or more of the n elements = 2^n.
Hi Mitch, can you give example with more rooms and more employee? I still not understand the formula here.

I used this formula, but more complicated:

NO one was assigned to the office: 3C0
just one employee was assigned to the office: 3C1
just 2 employees were assigned to the office: 3C2
all employees were assigned to the office: 3C3

Sum of all = 8
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