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Assign employees

by gibran » Thu May 15, 2008 7:28 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
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by punit.kaur.mba » Thu May 15, 2008 7:41 am
Either the 2question is incorrectly framed and should also consider the possibility that none of the employess are assigned to any office. That way the answer would be 5.

Else the answer should be 4 and i dont see that in the options!!

the 2 offices

Off 1, Off 2

No. of ways-

Off 1 - 3 , Off2 0
off1 - 2 , Off2 - 1
Off1 - 0, Off2 -3
Off1 - 1, Off2 - 2

Someone correct me if I m wrong!
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by aatech » Thu May 15, 2008 8:11 am
IMO D

Case 1 - One office has all 3 employees and second has none - this can happen in 2 ways

Case 2 - One office has 1 employee and another office has 2 - this can happen in 6 ways

So total 8
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by Magellan » Thu May 15, 2008 8:27 am
aatech wrote:Case 2 - One office has 1 employee and another office has 2 - this can happen in 6 ways
How is that possible? I see only two cases: 1 - 2 or 2 - 1
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by netigen » Thu May 15, 2008 11:38 am
0 | 3 -> 1 way to do this
1 | 2 -> 3 ways to do this
2 | 1 -> 3 ways to do this
3 | 0 -> 1 way to do this
-------------
Total = 8 (assuming each employee is a unique individual 8))
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by punit.kaur.mba » Thu May 15, 2008 11:54 am
Actually aatech is right. The answer is 8.

We are right in selecting cases but wrong in counting the number of arrangements in that case.

In second case

(2,1) and (1,2)

Each case inturn has 3 ways becos we are "arranging" people between 2 offices here.


So if E denotes Employee and O denotes office

Selecting E1 and E2 is not same as selecting E2 and E3.

So within the second case

1) 3 ways of arranging people in O1,O2
(a) O1 - E1 and O2-E2,E3
(b)O1-E2 and O2-E1,E3
(c)O1-E3 and O2-E1,E2

2) similar to above case but just reversed

(a) O1 - E2,E3, O2 - E1
(b) O1- E1,E3 O2-E2
(c)O1-E1,E2 O2- E3
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by Magellan » Thu May 15, 2008 12:11 pm
OK, I see. Difference between permutation and combination if I am not mistaken.
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by ektamatta » Thu May 15, 2008 6:33 pm
there is another way to solve this problem

2 office
3 employees

2^3

8 ways
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by lunarpower » Fri May 16, 2008 10:38 pm
ektamatta wrote:there is another way to solve this problem

2 office
3 employees

2^3

8 ways
that works, although it's certainly not self-explanatory enough as is.

here's a fuller explanation:
let's call the offices 1 and 2.
for each employee, there's a choice of being assigned either to office 1 or to office 2. furthermore, wherever the previous employees are assigned has no effect upon the assignments of the other employees, since double-/triple-packing and empty offices are both fine.

therefore:
3 employees
2 options for each employee
multiply the #s of options, because they're independent sequential choices: 2 x 2 x 2 = 8 total sets of choices.
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easy way

by umaa » Wed Jun 25, 2008 4:27 am
An easy way to find it. You have an equation to find this if there are no conditions. That is, n^r. which is 2^3=8.
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by gmatrant » Tue Aug 24, 2010 6:58 am
I know this is an old post.

But with reference to Ron's reply

Each employee actually has 3 options.
To choose Office 1 or office 2 or none of the offices
So wouldn't that make the choice
3*3*3 .

Ron - Please let me why this is wrong?
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by kapoor.divs » Tue Apr 26, 2011 3:51 am
gmatrant wrote:I know this is an old post.

But with reference to Ron's reply

Each employee actually has 3 options.
To choose Office 1 or office 2 or none of the offices
So wouldn't that make the choice
3*3*3 .

Ron - Please let me why this is wrong?
Hi gmatrant,

came across this thread.There is a 99% possibility that u must have already aced the gmat and I know the qs is almost a year old ,but still replying to this one.

each employee has to belong to an office and so he can choose either office 1 or office 2.
Not choosing office 1 will imply that he belongs to office 2 and vice versa.

Hope this makes things clear..

Cheers!
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by manishpal » Tue Sep 20, 2011 5:24 am
LUNARPOWER UR FACE IS LIKE SNOOPDOG..LOKKS LIKE A ROCKSTAR
lunarpower wrote:
ektamatta wrote:there is another way to solve this problem

2 office
3 employees

2^3

8 ways
that works, although it's certainly not self-explanatory enough as is.

here's a fuller explanation:
let's call the offices 1 and 2.
for each employee, there's a choice of being assigned either to office 1 or to office 2. furthermore, wherever the previous employees are assigned has no effect upon the assignments of the other employees, since double-/triple-packing and empty offices are both fine.

therefore:
3 employees
2 options for each employee
multiply the #s of options, because they're independent sequential choices: 2 x 2 x 2 = 8 total sets of choices.
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by manishpal » Tue Sep 20, 2011 5:27 am
A REPLY BY tictaktoe
REALLY EASY TO UNDERSTAND IN THIS WAY.

let x,y b office and A,B AND C be employyes

OFFICE X ------ OFFICE Y

ABC ----- nil
nil------- ABC
AB------- C
CA --------B
BC --------A
A ---------BC
C ---------AB
B ---------CA
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by talaangoshtari » Wed Jul 15, 2015 2:45 am
Why is 3^2 wrong? If we assume that we have 2 slots, for 2 offices. For each office we can assign 3 people, 3 × 3 = 9
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