Problem Solving

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?

I came up with the two different explanation, both reaching the same answer

1> total combinations = 3c0 + 3c1 + 3c2 + 3c3 = 8

2> Each employee can be assigned to two offices in 2 ways.
So three employees can be assigned in 2 x 2 x 2 = 8 ways.


OA-8

Which one is correct.

Thanks

gmat740 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?

I came up with the two different explanation, both reaching the same answer

1> total combinations = 3c0 + 3c1 + 3c2 + 3c3 = 8

2> Each employee can be assigned to two offices in 2 ways.
So three employees can be assigned in 2 x 2 x 2 = 8 ways.


OA-8

Which one is correct.

Thanks

I think that the second approach is the right one in this case.

I agree with the 2nd approach.

Cheers.

gmat740 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?

I came up with the two different explanation, both reaching the same answer

1> total combinations = 3c0 + 3c1 + 3c2 + 3c3 = 8

2> Each employee can be assigned to two offices in 2 ways.
So three employees can be assigned in 2 x 2 x 2 = 8 ways.


OA-8
Thanks

BUt I find a difficulty accepting the second approach...

First of all I can't understand why each employees can be assigned to two offices in 2 ways. I can be done three ways.

assign office 1, assign office 2, assign none.

so three. Then total ways 3+3+3 = 9 (for three employees)

But all offices can not be unassigned. In the above 9 cases one case is so.
so the no of ways=9-1=8


please comment...

I think you are confused between
1) some of the offices can be empty..

and

2) employees can be assigned to none of the offices


the question states that the office may have 0 employees...it does NOT state that the employee could be assigned to no office...if it stated that employee may or may not be assigned to the office then i think you would be correct in thinking of the 3 possibilities for each employee...

Cheers.

real2008 wrote:

gmat740 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?

I came up with the two different explanation, both reaching the same answer

1> total combinations = 3c0 + 3c1 + 3c2 + 3c3 = 8

2> Each employee can be assigned to two offices in 2 ways.
So three employees can be assigned in 2 x 2 x 2 = 8 ways.


OA-8
Thanks

BUt I find a difficulty accepting the second approach...

First of all I can't understand why each employees can be assigned to two offices in 2 ways. I can be done three ways.

assign office 1, assign office 2, assign none.

so three. Then total ways 3+3+3 = 9 (for three employees)

But all offices can not be unassigned. In the above 9 cases one case is so.
so the no of ways=9-1=8


please comment...