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3 employees, 2 offices

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3 employees, 2 offices

by alex.gellatly » Thu Jun 28, 2012 5:20 pm
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

Thanks
Last edited by alex.gellatly on Thu Jul 12, 2012 5:36 pm, edited 1 time in total.
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by eagleeye » Thu Jun 28, 2012 5:38 pm
alex.gellatly 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?
Hi Alex:

A couple of ways of doing this:

1. From the employees perspective.
Each employee has the option to go to either of the 2 offices. So number of ways of assigning employees to offices = the number of ways of each employee selecting an office without restriction = [spoiler]2*2*2 = 8
[/spoiler]
2. From the office filling perspective.

The two possibilities are:
a. One of the offices has 3 employees and the other 0. No. of ways of selecting which office is going to be filled = 2C1 = 2.
b. One of the offices has 2 employees and the other has 1.
b1. First let's select which office gets two employees. No of ways of doing this = 2C1 = 2.
b2. Second, let's select which of the two employees go to the office selected above. No. of ways
of doing so = 3C2 = 3.
So total number of ways of selecting an office and putting two people in it. = 2C1*3C2 = 2*3 = 6.

Hence total number of ways = [spoiler]2+6 = 8. [/spoiler]

D is correct.

Let me know if this helps :)
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by alex.gellatly » Wed Jul 11, 2012 12:20 am
eagleeye wrote:
alex.gellatly 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?
Hi Alex:

A couple of ways of doing this:

1. From the employees perspective.
Each employee has the option to go to either of the 2 offices. So number of ways of assigning employees to offices = the number of ways of each employee selecting an office without restriction = [spoiler]2*2*2 = 8
[/spoiler]
2. From the office filling perspective.

The two possibilities are:
a. One of the offices has 3 employees and the other 0. No. of ways of selecting which office is going to be filled = 2C1 = 2.
b. One of the offices has 2 employees and the other has 1.
b1. First let's select which office gets two employees. No of ways of doing this = 2C1 = 2.
b2. Second, let's select which of the two employees go to the office selected above. No. of ways
of doing so = 3C2 = 3.
So total number of ways of selecting an office and putting two people in it. = 2C1*3C2 = 2*3 = 6.

Hence total number of ways = [spoiler]2+6 = 8. [/spoiler]

D is correct.

Let me know if this helps :)
Yes, your explination is clear, thanks a lot. I do, however, have one quick question. Why is my logic incorrect? I just used the simple approach (3*2=6). For example, if there are 3 main dishes and 2 drink options, there are 6 total options right? How is this question different?

Thanks
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by shekhar.kataria » Thu Jul 12, 2012 11:38 am
HI Eagleeye

I think you made a mistake of not considering a point made in the Question. It says Non of the offices be empty. However, you solution counts methods when all 3 are assingned to a sinle company.

Therefore i think we have to subtract two such cases and then the answer would be 6.

Correct em if i am wrong in my understanding. Thanks

eagleeye wrote:
alex.gellatly 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?
Hi Alex:

A couple of ways of doing this:

1. From the employees perspective.
Each employee has the option to go to either of the 2 offices. So number of ways of assigning employees to offices = the number of ways of each employee selecting an office without restriction = [spoiler]2*2*2 = 8
[/spoiler]
2. From the office filling perspective.

The two possibilities are:
a. One of the offices has 3 employees and the other 0. No. of ways of selecting which office is going to be filled = 2C1 = 2.
b. One of the offices has 2 employees and the other has 1.
b1. First let's select which office gets two employees. No of ways of doing this = 2C1 = 2.
b2. Second, let's select which of the two employees go to the office selected above. No. of ways
of doing so = 3C2 = 3.
So total number of ways of selecting an office and putting two people in it. = 2C1*3C2 = 2*3 = 6.

Hence total number of ways = [spoiler]2+6 = 8. [/spoiler]

D is correct.

Let me know if this helps :)
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by eagleeye » Thu Jul 12, 2012 12:43 pm
alex.gellatly wrote:
eagleeye wrote:
alex.gellatly 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?
Hi Alex:

A couple of ways of doing this:

1. From the employees perspective.
Each employee has the option to go to either of the 2 offices. So number of ways of assigning employees to offices = the number of ways of each employee selecting an office without restriction = [spoiler]2*2*2 = 8
[/spoiler]
2. From the office filling perspective.

The two possibilities are:
a. One of the offices has 3 employees and the other 0. No. of ways of selecting which office is going to be filled = 2C1 = 2.
b. One of the offices has 2 employees and the other has 1.
b1. First let's select which office gets two employees. No of ways of doing this = 2C1 = 2.
b2. Second, let's select which of the two employees go to the office selected above. No. of ways
of doing so = 3C2 = 3.
So total number of ways of selecting an office and putting two people in it. = 2C1*3C2 = 2*3 = 6.

Hence total number of ways = [spoiler]2+6 = 8. [/spoiler]

D is correct.

Let me know if this helps :)
Yes, your explination is clear, thanks a lot. I do, however, have one quick question. Why is my logic incorrect? I just used the simple approach (3*2=6). For example, if there are 3 main dishes and 2 drink options, there are 6 total options right? How is this question different?

Thanks
Hi alex:
Your approach answered the following question. In how many ways can we select one drink and one meal from a selection. In this case, we always have a drink present. In the office employees question (where office is analogous to a drink and employee to a meal), we have the option that "some of the offices may be empty". If the question said, "none of the offices may be empty", 3*2 would be correct.

I hope this helps :)
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by eagleeye » Thu Jul 12, 2012 1:07 pm
shekhar.kataria wrote:HI Eagleeye

I think you made a mistake of not considering a point made in the Question. It says Non of the offices be empty. However, you solution counts methods when all 3 are assingned to a sinle company.

Therefore i think we have to subtract two such cases and then the answer would be 6.

Correct em if i am wrong in my understanding. Thanks

eagleeye wrote:
alex.gellatly 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?
Hi Alex:

A couple of ways of doing this:

1. From the employees perspective.
Each employee has the option to go to either of the 2 offices. So number of ways of assigning employees to offices = the number of ways of each employee selecting an office without restriction = [spoiler]2*2*2 = 8
[/spoiler]
2. From the office filling perspective.

The two possibilities are:
a. One of the offices has 3 employees and the other 0. No. of ways of selecting which office is going to be filled = 2C1 = 2.
b. One of the offices has 2 employees and the other has 1.
b1. First let's select which office gets two employees. No of ways of doing this = 2C1 = 2.
b2. Second, let's select which of the two employees go to the office selected above. No. of ways
of doing so = 3C2 = 3.
So total number of ways of selecting an office and putting two people in it. = 2C1*3C2 = 2*3 = 6.

Hence total number of ways = [spoiler]2+6 = 8. [/spoiler]

D is correct.

Let me know if this helps :)
.

Hi shekhar.kataria:
You didn't make a mistake, it's just a misinterpretation. The correct question reads "some of the offices can be empty". When Alex posted the question the "s" in some was missing, and it shows "ome of the offices". You interpreted it as "none of the offices". That's about it.
Last edited by eagleeye on Thu Jul 12, 2012 6:27 pm, edited 1 time in total.
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by alex.gellatly » Thu Jul 12, 2012 5:38 pm
SORRY!!! For the typo... it has been fixed.

@eagleeye: Thank you very much for your clear explanations.
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