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GMAT215
Pls post official solution :

this is how I worked :

Situation I : all 3 employees working:
# ways to choose 3 employees = 1
all three working of same project : 4 ways
all three working on different projects : 4 ways * 3 ways * 2 ways : 24 ways
2 working of same project & 1 working on different project : 4 ways * 3 ways : 12 ways

Total number of ways : 1*(4 + 24 + 12) = 40

Situation II : 2 out of 3 employees working:
# ways to choose 2 employees : 3 * 2 / 2 = 3
both working of same project : 4 ways
both working on different projects : 4 ways * 3 ways : 12 ways

Total number of ways : 3* (4 + 12) = 48

Situation III : 1 out of 3 employees working:
# ways to choose 1 employees : 3
working on any of the project : 4

Total number of ways : 3*(4) = 12

Situation IV : No employee working :
Total number of ways : 1


Sum of Situation I to IV : (40 + 48 + 12 + 1) = 101


I know its not in match with any of the possible solutions, pls help me to know where I have made mistake.

Hey GMAT215

I think you have misunderstood the question. I would request you to read it again and focus on the highlighted portion. :)

A certain company assigns projects to employees such that more than one or no employee can work on any project. In how many ways 3 employees will be assigned to 4 projects.

    The constraint in this question is that there could be a project in which more than one employee can work OR no one can work on that project!

    This however, does not mean that there could be a case where all 3 employees are working or 2 are working or 1 are working or 0 are working.

The constraint is on the project not on the people. We are explicitly saying that there 3 employees will be assigned to any of these 4 projects. We just need to find in how many ways, keeping in mind that there could be a project in which no one is working or more than one is working.

Keeping the above points in mind, can you try this question again?


Regards,
Saquib
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reveal explanation to this question please.
According to me 1 employee has 4 options either go in any of the three or do not go at all so 4 ways
so for all three employees 4*4*4 =64 ways
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Hey Everyone,

Official solution to the question has been posted.

Regards,
Ashutosh
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Solution



Given:
    • The company has to assign 3 employees to 4 projects
    • None or more than one employee can be assigned to any project.

To find:
    • The number of ways to assign 3 employees to 4 projects

Approach and Working:


Method-1)

Every employee can work in any of the 4 projects. Hence, 3 employees can work in 4*4*4=64 ways.

Method-2) By applying \(r^n\)

We have 3 employees or 3 objects that are to be arranged in 4 projects.

This is very similar to what we learnt in our article.

Thus,
    • n=3
    • r=4
And, total ways to arrange n objects into r things= \(4^3\)=64

Hence, the correct answer is option C.

Answer: C

Hi, your solution does not seem to be correct, as it includes cases in which a project can have only 1 employee assigned to it.

I solved it as below:
To Each project (p1, p2, p3, p4), employees ( e1, e2, e3) can be assigned in following way:

For p1,
Case 1: all 3 emp assigned = 1 way
Case 2: 2 out of 3 emp assign = 3C2 = 3 ways
Case 3: no emp assigned = 1 way
Total ways for p1 = 5

Hence no. of ways employees can be assigned to 4 projects = 5^4 = 625
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Hello everyone,

My question is if the given were written as 4 projects spread over 3 employees, would the formula be 3^4?

Thanks in advance
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Are we not considering case of one employee? But the question is asking to fing out eith none or more than one. I believe the question should include one .

A certain company assigns projects to employees such that one or more than one or no employee can work on any project. In how many ways 3 employees will be assigned to 4 projects?

But If we are considering the cases except one what should be the approach?

Posted from my mobile device
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Solution



Given:
    • The company has to assign 3 employees to 4 projects
    • None or more than one employee can be assigned to any project.

To find:
    • The number of ways to assign 3 employees to 4 projects

Approach and Working:


Method-1)

Every employee can work in any of the 4 projects. Hence, 3 employees can work in 4*4*4=64 ways.

Method-2) By applying \(r^n\)

We have 3 employees or 3 objects that are to be arranged in 4 projects.

This is very similar to what we learnt in our article.

Thus,
    • n=3
    • r=4
And, total ways to arrange n objects into r things= \(4^3\)=64

Hence, the correct answer is option C.

Answer: C

Doesn't it include the case where only one is working on a project ?
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The question is very easy but the English used is very confusing.
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Can anyone help me to understand the constraints in this question? Thanks in advance.
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GMAT215
Pls post official solution :

this is how I worked :

Situation I : all 3 employees working:
# ways to choose 3 employees = 1
all three working of same project : 4 ways
all three working on different projects : 4 ways * 3 ways * 2 ways : 24 ways
2 working of same project & 1 working on different project : 4 ways * 3 ways : 12 ways

Total number of ways : 1*(4 + 24 + 12) = 40

Situation II : 2 out of 3 employees working:
# ways to choose 2 employees : 3 * 2 / 2 = 3
both working of same project : 4 ways
both working on different projects : 4 ways * 3 ways : 12 ways

Total number of ways : 3* (4 + 12) = 48

Situation III : 1 out of 3 employees working:
# ways to choose 1 employees : 3
working on any of the project : 4

Total number of ways : 3*(4) = 12

Situation IV : No employee working :
Total number of ways : 1


Sum of Situation I to IV : (40 + 48 + 12 + 1) = 101


I know its not in match with any of the possible solutions, pls help me to know where I have made mistake.

Hey GMAT215

I think you have misunderstood the question. I would request you to read it again and focus on the highlighted portion. :)

A certain company assigns projects to employees such that more than one or no employee can work on any project. In how many ways 3 employees will be assigned to 4 projects.

    The constraint in this question is that there could be a project in which more than one employee can work OR no one can work on that project!

    This however, does not mean that there could be a case where all 3 employees are working or 2 are working or 1 are working or 0 are working.

The constraint is on the project not on the people. We are explicitly saying that there 3 employees will be assigned to any of these 4 projects. We just need to find in how many ways, keeping in mind that there could be a project in which no one is working or more than one is working.

Keeping the above points in mind, can you try this question again?


Regards,
Saquib
Quant Expert
e-GMAT

Hey saquib @e-gmat , thank you for your guidance. But, does not the constraint ""more than one employee can work OR no one can work on that project!"" refers that any project may have either 0 employees or more than 1 employees, but no project shall have 1 employee !!!
A strong doubt on the interpretation of the mentioned constraint.
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Given: A certain company assigns projects to employees such that more than one or no employee can work on any project.
Asked: In how many ways 3 employees will be assigned to 4 projects.

Number of ways 3 employees will be assigned to 4 projects = 4^3 = 64

IMO C
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I believe previous people have mentioned this but the way this question is stated is very confusing. Based on the way the question is written, we can't have ONLY 1 employee for a project, it must be either 0 or larger than 1.
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My solution, I don't know if it's acceptable even though I got the answer.

Case 1: 4P3 = 4 (3 people on a project)
Case 2: 4P2 = 12 (2 people on a project)
case 3: 4P1 = 24 (Each person per project)
case 4: 4P0 = 24 (Nobody on any project)
Total = 64.
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more than one or no employee can work on any project

4^3 considers cases in which a project can also have 1 person. But the above statement does not include this case. Isn't this ambiguous?­
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Whenever I come across a question on forum quiz which has unclear language, I instantly know it's gotta do with Egmat. GMAT club needs to block them or do something around this problem.
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