A certain company assigns employees to offices in such a way

The best way to remember this is :
(Decisions) ^ (Players)
For this problem - 2 decisions , 3 players : 2^3=8

Each of three employee can be assigned to either of the two offices, meaning that each employee has 2 choices --> 2*2*2=2^3=8.

Answer: D.

Let X, Y and Z be the 3 employees.
Let A and B be the 2 offices.

Take the task of assigning the employees and break it into stages.

Stage 1: Assign employee X to an office
There two options (office A or office B), so we can complete stage 1 in 2 ways

Stage 2: Assign employee Y to an office
There two options (office A or office B), so we can complete stage 2 in 2 ways

Stage 3: Assign employee Z to an office
There two options (office A or office B), so we can complete stage 3 in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus assign all employees to offices) in (2)(2)(2) ways (= 8 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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i am still not able to understand. Can you please explain in detail?

also please tell me where i went wrong.This was my logic.

No. of people
office 1: 0|0|0|1|1|1|2|2|3
office 2: 1|2|3|0|1|2|0|1|0

this gives me 9 possible combination

The fastest way to solve this problem is by using the formula, 2^n, where n stands for the number of elements, or, in this case, the number of employees. This formula is derived from adding the number of combinations from

What’s important with this problem is not to treat it as a probability problem. While on the surface it may seem similar to a typical combinations problem, using the combinations formula to solve the problem is cumbersome.

Instead, use the formula, 2^n, where n stands for the number of elements, or, in this case, the number of employees.

This formula is derived from adding the number of combinations whenever you can select any number greater than zero and less than or equal to n. For instance, here we could have chosen any of three employees for the first office. So instead of using 3C0 + 3C1 + 3C2 + 3C3, we can use 2^3.

This formula becomes especially useful for larger numbers. Imagine the question were:

How many ways can 6 employees go in two offices?

(A) 8
(B) 32
(C) 48
(D) 64
(E) 120

Following the method of finding each case would take too much time. By using 2^n, we 2^6 = 64. (D)

Going back to my original point: do not think of this as a typical probability problem, but one that uses the 2^n concept. The problems, while not really a probability problem, employ the 2^n formula.

Positive integer N is the product of three distinct primes. How many factors in N?

Ans: 2^3 – 1 (zero is not a factor so hence we subtract one from 2^n).


A multiple-choice test has five possible answer choices. Any number of answers can be correct. (e.g. A-B-D is possible answer, C-D, or all five). How many different possible answers?

Ans: 2^5 – 1 = 31 You can’t leave question blank (like the empty office) so therefore -1.


By understanding the concept behind a question, instead of grouping a question under one general category, you should be able to solve problems more quickly.

Thank you. Then, if the company assigns employees to offices in such a way that if the offices can not be empty and more than one employee can be assigned to an office. And we have 5 employees and 3 rooms, the answer would be:

120?
I mean, 5!

No. It would be 3^5 minus restriction.

For example, for 5 employees and 2 offices it would be 2^5 - 2 ({5-0} and {0-5}).

ok, can someone tell me what's wrong with my thinking..
1st office can have any 3 employees.. therefore 3 options,
2nd office can also have any of 3 employees hence again 3 options
so it should be 3*3=9

i think the logic is similar to the way Bunuel did..the only difference is in that case we had 2 choices for each employee therefore it was 2*2*2=8.. but why is the answer different in both cases?

We are distributing employees to the offices not vise-versa.

We need to determine in how many ways the company can assign 3 employees to 2 different offices when some of the offices can be empty and more than one employee can be assigned to an office.

Since there are 3 people and 2 offices, we have 3 options for each office. Thus, the employees can be organized in 2^3 = 8 possible ways.

Alternative solution:

If you have trouble understanding why there should be 2^3 = 8 possible ways to assign 3 employees in 2 different offices, we can list all the possible ways one can assign 3 employees (say A, B and C) to 2 different offices (Office 1 and Office 2).

1) Office 1: A, B, C and Office 2: no one

2) Office 1: A, B and Office 2: C

3) Office 1: A, C and Office 2: B

4) Office 1: B, C and Office 2: A

5) Office 1: A and Office 2: B, C

6) Office 1: B and Office 2: A ,C

7) Office 1: C and Office 2: A, B

8) Office 1: no one and Office 2: A, B, and C

As we can see, there are 8 ways to assign 3 employees to 2 different offices.

Answer: D

Responding to a pm:

No, there is a problem in the logic used. There are 4 ways of distributing employees to the 2 offices. The order in which we distribute them in immaterial. So if we have 1 in office 1 and 2 people in office 2, it doesn't matter whether you put the 1 person in first or the 2 people in first. The final distribution is the same hence the logic of multiplying by 2 is nor correct.

Note that when you put 1 person in office 1 and 2 people in office 2, there are 3 distinct ways of doing it since the people are distinct (say A, B and C)
So A in office 1 and B, C in office 2 is different from B in office 1 and A, C in office 2.

So there is 1 way or 0 in office 1 and 3 in office 2.
3 ways of putting 1 in office 1 and 2 in office 2.
3 ways of putting 2 in office 1 and 1 in office 2.
and 1 way of putting 3 in office 1 and 0 in office 2.

That is how you get 1 + 3 + 3 + 1 = 8 cases.

• Approach using \(2^3 \)already discussed. So giving another approach to this.

Given

• 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.

To Find

• The number of ways can the company assign 3 employees to 2 different offices.

Approach and Working Out

• There are 2 offices so an employee has 3 choices.

o Office 1, office 2, or nowhere.

• So 3 employees will have 3 × 3 = 9 choices.

o However, for a particular scenario, none of the employees will be assigned which is forbidden as the question suggests some of the offices can be empty.

• Answer = 9 – 1 = 8.

Correct Answer: Option D

Bunuel how would the answer to this question change if it said no more than one employee can be assigned to each office space ??
3*2= 6 ??

In this case the answer would be 0. You cannot assign 3 employees to 2 different offices, so that no more than one employee can be assigned to each office. Remember, you should assign ALL 3 employees to either of the offices. For example, if Tom is assigned to office #1 and Mary is assigned to office #2, then Kate must be assigned to either #1 or #2 and this will violate the restriction you are bringing.

Bunuel KarishmaB

Why can't we do employee distributiion to the office?

So 3 employee can be distributed to both office: 3*3=9 and then imply restriction and remove the case in which no employee is distributed to any office, so there will be 2 cases (no employee to office 1 and ofc2)
9-2 = 7
By this logic, i get 7 instead of 8, how to solve using this logic then?

I am not sure what logic you are using here. How did you get 3*3? What are the 3 options?
I have 3 employees so an office can get an employee in 4 ways - 0, 1, 2 or 3
All of these are allowed. But once number of employees are chosen for one office, they are fixed for the other. If I put 1 employee in office 1, automatically office 2 gets the leftover 2 employees. Also, we will need to count the number of ways in which 1 employee can be given to office 1. It can happen in 3 ways. Similarly I can give 2 employees to office 1 in 3 ways. Also there is only 1 way of giving 0 employees or 3 employees to office 1.
So overall, we again get 8 ways.

It is best to use 2*2*2 method.

its clearly given that offices can be zero why are qe not counting that