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

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

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...