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Bunuel
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A certain company assigns employees to offices in such a way 13 Nov 2024, 06:56
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Bunuel
sagarsabnis
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

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
First of all, you should assign all three employees to one of the two offices. You can have the following scenarios for the number of people in each office:

ABCD
Office 1 0123
Office 23210

In scenarios (A) and (D), there is only one way to assign the three people. However, in (B) and (C), there are 3 possible cases each:

Let’s say the employees are Tom, Mary, and Kate. In scenario (B): Tom could be in office #1, with Mary and Kate in office #2; or Mary could be in office #1, with Tom and Kate in office #2; or Kate could be in office #1, with Tom and Mary in office #2. So, there are 3 cases for (B) and the same for (C). Adding them up: (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. This gives a total number of combinations (assignments) of 2 * 2 * 2 = 2^3 = 8.

Hope it's clear.

its clearly given that offices can be zero why are qe not counting that
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The solution already considers all cases, including those where one office has zero employees. That’s why we have scenarios (A) and (D):

  • In (A), Office 1 has 0 people, and Office 2 has 3 people.
  • In (D), Office 1 has 3 people, and Office 2 has 0 people.

All possible distributions are accounted for across scenarios (A), (B), (C), and (D), so there's nothing missing.
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A certain company assigns employees to offices in such a way 12 Jan 2025, 11:21
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sagarsabnis
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
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either one office will be open or both office will be opened.

if 1 office is open, then we have 2 choices of office.

if both office is open, we can have 6 different arrangements. so total 8
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A certain company assigns employees to offices in such a way 24 Jan 2025, 20:15
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There are 3 people and they can all be assigned to either of 2 offices = 3 x 3 = 9. However, one of these options includes the possibility that all 3 are in both offices. So; (3 x 3) - 1 = 8 outcomes

Bunuel
This got me the right answer, but is the logic sound?
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A certain company assigns employees to offices in such a way 28 Jan 2025, 04:52
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they can be assigned to any office so 3*3 but one of these outcomes will include the fact that all 3 are in both offices so -1

(3 * 3) - 1

Bunuel chetan2u

does this work? It got me the answer but idk if the logic is sound
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A certain company assigns employees to offices in such a way 10 Feb 2025, 05:18
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2^8 = 256 (and not 64), kindly correct this in your example. Magoosh Test Prep

ChrisLele
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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A certain company assigns employees to offices in such a way 14 Apr 2025, 03:49
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Why the (n+r-1)C(r-1) formula doesn't work here?
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A certain company assigns employees to offices in such a way 14 Apr 2025, 03:57
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Why the (n+r-1)C(r-1) formula doesn't work here?
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Share your thoughts on why do you think it should work :)
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A certain company assigns employees to offices in such a way 14 Apr 2025, 04:01
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I don't know, that's why I'm asking :)

Isn't it similar scenario where 5 apples are distributed among 4 children and some children may get zero apples.

Krunaal
JT2916
Why the (n+r-1)C(r-1) formula doesn't work here?
Share your thoughts on why do you think it should work :)
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A certain company assigns employees to offices in such a way 14 Apr 2025, 04:18
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JT2916
I don't know, that's why I'm asking :)

Isn't it similar scenario where 5 apples are distributed among 4 children and some children may get zero apples.

Krunaal
JT2916
Why the (n+r-1)C(r-1) formula doesn't work here?
Share your thoughts on why do you think it should work :)
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Ok, so you thought it should work because you misunderstood it for a different type of problem - No this isn't a similar scenario, the 5 apples are identical there, here you have 3 different employees, and two different offices.

Check out this solution on first page of this problem: https://gmatclub.com/forum/a-certain-co ... ml#p671810

Here's a link to theory for different types of Combinatorics problems, and how to tackle each type: https://gmatclub.com/forum/combinatoric ... 06266.html

Hope it helps.
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A certain company assigns employees to offices in such a way 14 Apr 2025, 04:23
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Ah Okay, thanks for the links.
This seems insightful: Combinatorics problems :)

Thanks!

Krunaal
JT2916
I don't know, that's why I'm asking :)

Isn't it similar scenario where 5 apples are distributed among 4 children and some children may get zero apples.

Ok, so you thought it should work because you misunderstood it for a different type of problem - No this isn't a similar scenario, the 5 apples are identical there, here you have 3 different employees, and two different offices.

Check out this solution on first page of this problem: https://gmatclub.com/forum/a-certain-co ... ml#p671810

Here's a link to theory for different types of Combinatorics problems, and how to tackle each type: https://gmatclub.com/forum/combinatoric ... 06266.html

Hope it helps.
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A certain company assigns employees to offices in such a way 24 Jun 2025, 07:54
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Just a small doubt - how do we know what number to raise to which power? If we looked at the problem in a different way, we could have chosen to do 3 x 3 (the number of employees that can be assigned per office). Is this just because of the fact that the question says that one office can go empty, i.e. we do not have to assign an employee to the different offices, but each employee has to be assigned an office?
Bunuel
sagarsabnis
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
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D

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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A certain company assigns employees to offices in such a way 24 Jun 2025, 09:29
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Just a small doubt - how do we know what number to raise to which power? If we looked at the problem in a different way, we could have chosen to do 3 x 3 (the number of employees that can be assigned per office). Is this just because of the fact that the question says that one office can go empty, i.e. we do not have to assign an employee to the different offices, but each employee has to be assigned an office?
Bunuel
sagarsabnis
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
Show Spoiler
D

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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Because the question is about assigning 3 employees to 2 offices, each employee must be assigned to one office. So each employee has 2 choices, giving 2^3 = 8 ways.

If you flipped it and said each office has 3 choices, you'd be assigning multiple offices to the same employee, which makes no sense, since each employee can only go to one office.
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A certain company assigns employees to offices in such a way 25 Jun 2025, 00:08
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Thanks for the explanation! :)
Bunuel
kabirgandhi
Just a small doubt - how do we know what number to raise to which power? If we looked at the problem in a different way, we could have chosen to do 3 x 3 (the number of employees that can be assigned per office). Is this just because of the fact that the question says that one office can go empty, i.e. we do not have to assign an employee to the different offices, but each employee has to be assigned an office?


Because the question is about assigning 3 employees to 2 offices, each employee must be assigned to one office. So each employee has 2 choices, giving 2^3 = 8 ways.

If you flipped it and said each office has 3 choices, you'd be assigning multiple offices to the same employee, which makes no sense, since each employee can only go to one office.
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