A courier company can assign its employees to its offices in such a wa

Question

A courier company can assign its employees to its offices in such a way that at one or more of the offices can be assigned no employee to all the employees. In how many ways can the company assign four employees to two different offices?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 16

(A) 6
(B) 8
(C) 10
(D) 12
(E) 16

AvidDreamer09 · Answer

how many ways can the company assign four employees to two different offices??

A has 2 choices office 1 or office 2
B has 2 choices office 1 or office 2
C has 2 choices office 1 or office 2
D has 2 choices office 1 or office 2

Since this is a dependent Event... you would have to multiply 
 2*2*2*2=16

fskilnik · Answer

Hi,  Bunuel  ,

I suggest changing the wording (perhaps) to this:

Reason: the way originally presented, I thought we could let both two offices empty (eventually assigning the employees to other offices). 
AvidDreamer09´s solution is coherent to the modification suggested.

Regards,
Fabio.

AvidDreamer09 · Answer

I'm not sure if my thinking is right. I thought that too. I thought both the offices could be empty but I reasoned 3^4 was not part of the answer choices and well it would be a pointless exercise for the management to not allocate any employee to neither of the office. So I considered 2^4

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fskilnik · Answer

My answer would be 3^4, too.

Rethinking about the original question stem, now I understand the issue better. The "problem" is that the first sentence does not take into account there are only 2 offices (yet)...

I guess the wording should be changed. My (better) suggestion is the following:

Regards,
Fabio.

shaarang · Answer

I don't know if my reasoning is correct here, I'd appreciate the critique.

I solved this problem just like the chocolate distribution problem you see here:  https://gmatclub.com/forum/in-how-many-ways-5-different-chocolates-be-distributed-to-4-children-231187.html

1 room could go to: 1 emp, 2 emp, 3 emp or 4 emp -- 4 ways

2 rooms, so 4*4= 16

KarishmaB · Answer

This is how I read this question:

The company has to assign four employees to two different offices. There is no problem if an office has "no employees" or "all employees" or "anything in between".
But the employees HAVE to be assigned to an office.

So each employee has 2 options. 
So for 4 employees, we have 2*2*2*2 = 16 options

fskilnik · Answer

Hi, shaarang.

In AvidDreamer09´s solution (he certainly meant "INdependent", but I prefer the Fundamental Principle of Counting to justify those multiplications) 
and also in Karishma´s solution there is an implicit restriction that any given employee must be assign to exactly one of the two offices.

In your suggestion, it would be 
> possible that some employees were assigned to both offices
> possible that some employees were not assigned to any of the 2 offices 
> impossible to have an "empty" office (without employees assign to it)

I hope this is "the critique" you asked.

Regards,
Fabio.

Mridul94 · Answer

Why cant we use the formula for dividing n identical items between r persons here? n+r-1Cr-1

kiranteja93 · Answer

Hi all,
I solved it in the following way:
Two offices for four employees and each office can have 0 to four employees.
Now Imagine there are two columns O1 and O2 representing the offices and there are the the following possibilities for allocation.Also note that in each office order doesn't matter ; all that counts is who all I am selecting for first office and thereby implying that the selection for the other office is already done.In other words if I select for O1 a certain set of people then the remaining people will automatically be selected to O2.Now let's consider the possibilities.
1)O1 : 0 people,O2 : 4 people
Number of ways of doing this is only one.note that order doesn't matter 
2)O1 : 1 person O2: remaining 3 people
First let's select for O1 there are 4C1 ways and each way O2 people automatically gets selected
3)O1: 2 people O2 remaining 2 people
4C2 ways notice O2 automatically gets selected.

In this ways all cases are ored as 1 + 4C1 + 4C2 + 4C3 + 4C4 + 1 = 16 ways

Notice in each case I am considering the following for O 1 and O2 : 0,4. 1,3. 2,2. 3,1. 4,0.

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Aannie · Answer

Can someone explain why the following approach is wrong :

2 offices and 4 employees and each of the offices can have 0 employee assigned or even all employees assigned

Office 1. Office 2 
0 emp. 0 emp
1 emp. 3 emp
2 emp. 2 emp
3 emp. 1 emp
4 emp. 0 emp
0 emp. 4 emp

So total 6 ways .

Thanks in advance

MathRevolution · Answer

Total employees = 4

Every employee has 2 options: Either assigned to anyone or both offices or Not assigned.

=> 2 * 2 * 2 * 2 = 16

Answer E

StacyArko · Answer

Let’s start by grouping the options. 
1. We have two ways of assigning 4 employees to one office and 0 to the second office
2. We have two ways of assigning 3 employees to one office and 1 to the second office
3. We have one way of assigning two employees to office 1 and two to office 2.

For (1); (4C4 X 2) = 2
For (2); (4C1X 3C3 X 2) = 8
For (3); (4C2 X 2C2) = 6

Total = 2+8+6 = 16

Posted from my mobile device

bumpbot · Answer

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A courier company can assign its employees to its offices in such a wa

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