ketkoag 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?
A. 5
B. 6
C. 7
D. 8
E. 9

OA: D
I got the right answer by counting all the possible combinations.
please tell me if there is any better way to solve this..

We can break the task into 3 stages and apply the Fundamental Counting Principle.
Stage 1: Place the first employee. Two offices --> 2 ways to place the employee
Stage 2: Place the second employee. Two offices --> 2 ways to place the employee
Stage 3: Place the third employee. Two offices --> 2 ways to place the employee

The total number of ways to accomplish all three stages = 2x2x2 = 8

ketkoag wrote:thanks, i think counting is the best way to solve this question.

If counting works for you on this problem, that's great. For some problems, counting can be the best/fastest method and test takers should keep this in mind.
At the same time, you need to have some counting techniques in your repertoire when the numbers get too large. For example, this question could have included 8 employees and 5 offices, in which case listing and counting each possibility would be difficult.