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permutations and combination

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by naaga » Tue Mar 03, 2009 6:06 am
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

friends please explain , I am getting only 4 ways but the OA is A
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by kamu » Tue Mar 03, 2009 6:11 am
Hi,
there are two offices and three employees.
so three employees can be assigned to one room(say A) in 1 way
in the same way three employees can be assigned to one room(say B) in 1 way
and
two employees can be put in one room(A) and one employee in another room(B) in 3C1*2C2 ways=3
and
two employees can be put in one room(B) and one employee in another room(A) in 3C1*2C2 ways=3
total is 1+1+3+3=8 ways.
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rep

by naaga » Tue Mar 03, 2009 6:18 am
but OA is only 5 ways.

where we are going wrong ? I also got 8 ways.
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by krisraam » Tue Mar 03, 2009 6:31 am
Each employee has 2 options

Total number of ways = 2^3 = 8

Thanks
raama
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Re: rep

by kamu » Tue Mar 03, 2009 6:38 am
naaga wrote:but OA is only 5 ways.

where we are going wrong ? I also got 8 ways.
Use search option. Saves time.

OA is incorrect for this btw.
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