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Why not E?

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Why not E?

by massi2884 » Tue Apr 24, 2012 5:42 am
[spoiler]Can you explain why do we have to assign 2 possible offices to 3 employees (2*2*2=8) like in OA D and not 3 possible employees to 2 offices (3*3=9) like in wrong answer E? Thanks.[/spoiler]

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

[spoiler]Can you explain why do we have to assign 2 possible offices to 3 employees (2*2*2=8) like in OA D and not 3 possible employees to 2 offices (3*3=9) like in wrong answer E? Thanks.[/spoiler]
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by shantanu86 » Tue Apr 24, 2012 6:00 am
IMO its D(8) and not 9. Here is the reasoning-
---------------------------------
Off-1 Off-2 Combinations
---------------------------------
3 0 1
2 1 3C1 = 3 (Choose 1 employee in 3 to occupy Off-2)
1 2 3C2 = 3
0 3 1
0 0 0 , this is invalid as all employees must be assigned an office.
---------------------------------

Therefore, total number of combinations is 8.

Hope it helps!!
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by Anurag@Gurome » Tue Apr 24, 2012 6:14 am
massi2884 wrote:[spoiler]Can you explain why do we have to assign 2 possible offices to 3 employees (2*2*2=8) like in OA D and not 3 possible employees to 2 offices (3*3=9) like in wrong answer E? Thanks.[/spoiler]

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

[spoiler]Can you explain why do we have to assign 2 possible offices to 3 employees (2*2*2=8) like in OA D and not 3 possible employees to 2 offices (3*3=9) like in wrong answer E? Thanks.[/spoiler]
Each of three employee can be assigned to either of offices, meaning that each has 2 choices implies # of ways the company can assign 3 employees to 2 different offices = 2 * 2 * 2 = 2^3 = 8

The correct answer is D.
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by [email protected] » Tue Apr 24, 2012 10:42 pm
IMO its D(8) and not 9. Here is the reasoning-
---------------------------------
Off-1 Off-2 Combinations
---------------------------------
3 0 1
2 1 3C1 = 3 (Choose 1 employee in 3 to occupy Off-2)
1 2 3C2 = 3
0 3 1
0 0 0 , this is invalid as all employees must be assigned an office.
---------------------------------

Therefore, total number of combinations is 8.

A wonderful reasoning given...

I almost got the 4 options listed but could not somehow use the combinations formula for the 3 employees. So initially I got the answer as 4 which is not in the list...

Yes this is the correct method as it also gives you the overall picture. I liked it...

{a bit weak in probability}
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by ronnie1985 » Wed Apr 25, 2012 10:10 am
The question should have explicitly stated that both the offices must not be vacant.
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by shantanu86 » Wed Apr 25, 2012 10:23 pm
ronnie1985 wrote:The question should have explicitly stated that both the offices must not be vacant.
Hi ronnie1985,

"In how many ways can the company assign 3 employees to 2 different offices? "

IMHO above lines clearly state that every employee must be assigned to one of the offices under the constrains set.

To cross check-
If it was allowed to not assign some/all of employees, the number of arrangements will become much more than 9, the max option in list.

Hope it helps!!
If you feel like it, hit thanks :)
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