# Sturggling to arrive at the correct answer (Permutations)

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• May 2nd 2010, 02:36 PM
director
Sturggling to arrive at the correct answer (Permutations)
The problem is copied straight from the book:
Untitled Document

Section: Permutations
Quote:

11) There are n applicants for the director of computing. The applicants are interviewed
independently by each member of the three-person search committee
and ranked from 1 to n. A candidate will be hired if he or she is ranked first
by at least two of the three interviewers. Find the probability that a candidate
will be accepted if the members of the committee really have no ability at all
to judge the candidates and just rank the candidates randomly. In particular,
compare this probability for the case of three candidates and the case of ten
candidates.

The solution provided: (3n-2)/n^3

My approach:
There are 3 committee members A,B,C.
Each has his own ranking of candidates which can arranged in n! ways.
We want our applicant X in a certain position (1st place) in 2/3 rankings.
There are (n-1)! ways to arrange the other applicants in each ranking.

The "good" rankings (in order from 1st to last place):
A: X * * * * * * n
B: X * * * * * * n
C: * * * * * * * n

A: X * * * * * * n
B: * * * * * * * n
C: X * * * * * * n

A: * * * * * * * n
B: X * * * * * * n
C: X * * * * * * n

The number of ranking arrangements that make X the top applicant is:

3(n!(n-1)!(n-1)!)

The total number of ways to arrange the rankings is:

n! n! n!

Probability = good rankings/total rankings

3 (n! (n-1)! (n-1)!)
________________
(n! n! n!)

Simplifying:

3 n! (n-1)! (n-1)!
____________________
n(n-1)! n(n-1)! n(n-1)!

3 n(n-1)!
_________
n n n (n-1)!

= 3n / n^3

What am I doing wrong? Am I not accounting for repetition? This is killing me. (Lipssealed)
Appreciate any help.
• May 2nd 2010, 02:54 PM
Plato
The key to this in the fact of independence: they interview independently. It is binomional.
Applicant A has the probability of $\frac{1}{n}$ of being first, independently, on any given list.
A gets the job if A is first on two or three lists.
$\binom{3}{2}\left(\frac{1}{n}\right)^2\left(1-\frac{1}{n}\right)+ \left(\frac{1}{n}\right)^3$
• May 4th 2010, 08:53 AM
director
Thank you for the reply. I understand the answer now. Close-minded thinking on my part. :)