Combinatorics: Choosing Job Applicants

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Homework Statement


There are eight applicants for a job, and three different judges who each rank the three applicants. Applicants are chosen if an donly if they appear in the top three in all three rankings.

a) How many ways can the three judges produce their three rankings?

b) What is the probability tha tMr. Dickens, one of the applicants, being chosen in a random set of three rankings?



Homework Equations



P(n,r) and C(n,r)

The Attempt at a Solution



a) = 3 * C(8,3)

b) I am feeling really uneasy about my solution, but I think that if a judge picks Mr.
Dickens, then he has C(7,2) ways to pick the remaining applicants. So the probability
of Mr. Dickens being chosen in a random set of three rankings is [3*C(7,2)]/[3*C(8,3)]

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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a) is wrong. Each judge can pick the 9 candidates in any order. How many ways are there to pick them for each judge?
 
  • #3
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a) is wrong. Each judge can pick the 9 candidates in any order. How many ways are there to pick them for each judge?

Sorry, but I guess I'm really unclear now. I'm not sure if your number '9' was a typo or not. But if it was a typo, then I am not sure where the number '9' is coming from.

If it was a typo, and was supposed to be '8', then since I started this thread I have become burdened by wondering if the answer is C(8,3)^3. But since one judge can pick eight applicants in one way, that seems independent of what judges two and three do. That is why I initially added them together to get 3*C(8,3).

I'm honestly just not sure which one to use.
 
  • #4
Dick
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Sorry, but I guess I'm really unclear now. I'm not sure if your number '9' was a typo or not. But if it was a typo, then I am not sure where the number '9' is coming from.

If it was a typo, and was supposed to be '8', then since I started this thread I have become burdened by wondering if the answer is C(8,3)^3. But since one judge can pick eight applicants in one way, that seems independent of what judges two and three do. That is why I initially added them together to get 3*C(8,3).

I'm honestly just not sure which one to use.

Sure, I meant 8. Sorry. But each judge picks 8 candidates in a specific order. Isn't that P(8,8)?
 
  • #5
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Sure, I meant 8. Sorry. But each judge picks 8 candidates in a specific order. Isn't that P(8,8)?

That makes sense to me. But since 'a' is asking me "How many ways can the three judges produce their three rankings", I am unsure what P(8,8) buys me. So that is why I am saying it is either 3*C(8,3) or C(8,3)^3, since any one of the eight applicants can wind up in any one of the top three spaces.

So I'm totally unclear then how P(8,8) helps me out.
 
  • #6
Dick
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That makes sense to me. But since 'a' is asking me "How many ways can the three judges produce their three rankings", I am unsure what P(8,8) buys me. So that is why I am saying it is either 3*C(8,3) or C(8,3)^3, since any one of the eight applicants can wind up in any one of the top three spaces.

So I'm totally unclear then how P(8,8) helps me out.

The total number of ways the judges can produce their rankings doesn't depend on who is in the top 3.
 
  • #7
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The total number of ways the judges can produce their rankings doesn't depend on who is in the top 3.

Okay, I think I may be getting what you are saying. But I think I still have a gap in my understanding. You're saying that the total number of rankings of the eight applicants = 8!, or P(8,8). But it seems that there can be repetitive ways to have the top three rankings.

If the names of the applicants are 1,2,3,4,5,6,7,8 then,

1,2,3,4,5,6,7,8 and 1,2,3,4,6,5,8,7 are different rankings of the eight applicants, but they both have 1,2,3 in the top three positions. Let us call those rankings x and y. So since question 'a' is asking "How many ways can the three judges produce their three rankings?", if I say the answer is 8!, then aren't I counting x and y twice since the top three rankings are identical?
 
  • #8
Dick
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Okay, I think I may be getting what you are saying. But I think I still have a gap in my understanding. You're saying that the total number of rankings of the eight applicants = 8!, or P(8,8). But it seems that there can be repetitive ways to have the top three rankings.

If the names of the applicants are 1,2,3,4,5,6,7,8 then,

1,2,3,4,5,6,7,8 and 1,2,3,4,6,5,8,7 are different rankings of the eight applicants, but they both have 1,2,3 in the top three positions. Let us call those rankings x and y. So since question 'a' is asking "How many ways can the three judges produce their three rankings?", if I say the answer is 8!, then aren't I counting x and y twice since the top three rankings are identical?

I guess it depends on how you read 'three' in the problem. If you read it as how many ways can a judge rank the top three then sure, it's P(8,3). It took it to mean how many ways can a judge rank all eight candidates, then a single judge counts for P(8,8). Cubed, of course, since there are three judges.
 
  • #9
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I guess it depends on how you read 'three' in the problem. If you read it as how many ways can a judge rank the top three then sure, it's P(8,3). It took it to mean how many ways can a judge rank all eight candidates, then a single judge counts for P(8,8). Cubed, of course, since there are three judges.

Okay, so presuming the question is asking about how the judges can rank the top three, then my solution for 'b' is that I go ahead and automatically put Mr. Dickens in the top three rankings of all judges, and that leaves me with P(7,2)^3.

Thus, the probability of Mr. Jenkins being chosen is P(7,2)^3/P(8,3)^3. Does that all look about right?
 
  • #10
Dick
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Okay, so presuming the question is asking about how the judges can rank the top three, then my solution for 'b' is that I go ahead and automatically put Mr. Dickens in the top three rankings of all judges, and that leaves me with P(7,2)^3.

Thus, the probability of Mr. Jenkins being chosen is P(7,2)^3/P(8,3)^3. Does that all look about right?

Not quite. Dickens could be first, second or third.
 
  • #11
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Not quite. Dickens could be first, second or third.

So if I place him in first place, that would be P(7,2). I'm getting mixed up for the second and third places. If I place Mr. Dickens in 2nd place, then is it still P(7,2), and likewise for third place. So why wouldn't the numberator be P(7,2)^3? Plus, that is just the choices for one judge. So would I have to cube P(7,2)^3 to get [P(7,2)^3]^3
 
  • #12
Dick
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So if I place him in first place, that would be P(7,2). I'm getting mixed up for the second and third places. If I place Mr. Dickens in 2nd place, then is it still P(7,2), and likewise for third place. So why wouldn't the numberator be P(7,2)^3? Plus, that is just the choices for one judge. So would I have to cube P(7,2)^3 to get [P(7,2)^3]^3

If you consider a single judges choices, if Dickens is first there are P(7,2) choices for the other slots. Ditto for second and third. They are all different cases. I make that 3*P(7,2).
 
  • #13
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Thank you for your help with this problem.
 

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