First of all you should assign ALL 3 employees to either of the offices. You can have the following scenarios:

No. of people
***********A|B|C|D|
office 1: 0|1|2|3|
office 2: 3|2|1|0|

In scenario (A) and (D) there is only one way to assign three people. But in (B) and (C) there will be 3 cases in each:

Let's say there are 3 employees: Tom, Mary and Kate. In (B): Tom can be in office #1 and Mary/Kate in #2 OR Mary can be in #1 and Tom/Kate in #2 OR Kate in #1 and Tom/Mary in #2. Total 3 cases for (B). The same for (C). (A)+(B)+(C)+(D)=1+3+3+1=8.

The way I solved this was different:

Each of the three employees, Tom, Mary and Kate, has two choices office #1 or office #2. Hence total # of combinations (assignments) is 2*2*2=2^3=8.

Hope it's clear.
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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.
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For each one of the 3 employees, there are two choices. He can be allotted to any one of the two offices. Hence total number ways will be 2 * 2* 2 = 8 ways
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Think in this way:
There is no restriction on the offices i.e. they can be vacant, they can accommodate all 3 employees etc. But there is a restriction on the employees i.e. each one of them must get an office.

Employee 1 can get an office in 2 ways - office A or office B
Employee 2 can get an office in 2 ways - office A or office B
Employee 3 can get an office in 2 ways - office A or office B
All three can be allotted offices in 2*2*2 = 8 ways
This takes care of all cases.
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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 8 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 2^8 = 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.
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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}).
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Applying combinations I think would be in this way:
\(C^4_1 * C^2_1 = 4*2 = 8\)

We are distributing employees to the offices not vise-versa.
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with 2 offices and 3 employees,
there are 3 ways to have one's own office,
3 ways to share an office with one other, and
2 ways to share an office with two others, or
8 ways total

Another way to put it:

There are 3 employees in total, so the max number of people in one office is 3. And the min is 0.

Ways to distribute 3 employees (E) in 2 offices.

Office 1:

E E E
E E _
E _ _
_ _ _ (empty)

Office 2: (Same as office 1.)

E E E
E E _
E _ _
_ _ _ (empty)


Since there is no distinction between each employee for this particular question, there are 4 ways to have 3 employees fill an office. Since there are two offices, there are:
4 + 4 = 8 ways to assign 3 employees to 2 offices


Would this approach be correct? VeritasPrepKarishma