Difficult one for me

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Difficult one for me

by abhi.genx7 » Fri Oct 01, 2010 11:17 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
Source: — Problem Solving |

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by shovan85 » Fri Oct 01, 2010 11:30 am
IMO D

2 offices 3 employees to be put

1st O - 0 E 2nd O - 3 E => this can be done in 2 ways
1st O - 1 E 2nd O - 2 E => this can be done in 3 ways
1st O - 2 E 2nd O - 1 E => 3 ways.

total 8 :)

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by Brian@VeritasPrep » Fri Oct 01, 2010 11:44 am
Great work, Shovan - I agree, and you're on fire today!

I was about to post pretty much the same thing, and add the strategic tip that:

When dealing with arrangement problems that use small numbers (like 2, 3, 4), if you don't immediately know the permutations/combinations formula to use you're usually much, much faster if you just write out the possibilities. These problems can be a lot more conceptual than mechanical - I've seen people waste minutes trying to figure out how to apply the formula to a problem like this when you can draw up a quick chart and be done in less than a minute:


Office 1 Office 2
ABC none
none ABC
A BC
BC A (note that our second and fourth entries are just the opposites of the first and third)
B AC
opposite of above
C AB
opposite of above

By jotting these down, you can either exhaust your options at 8 fairly quickly, or determine a system that works for better organizing them (there are two offices, so determine how many ways to arrange the first office and then multiply by 2 for "it can be flipped")

When these problems use smaller numbers, it's pretty likely that they're testing you conceptually, so use that kind of reasoning and don't worry as much about finding formulas.
Brian Galvin
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Chief Academic Officer
Veritas Prep

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by TheCloakedMonk » Fri Oct 01, 2010 12:19 pm
A:BC
BC:A
B:AC
AC:B
C:AB
AB:C
ABC:NONE
NONE:ABC
_________
TOTAL = 8
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