The problem is copied straight from the book:

Untitled Document

Section: Permutations

The solution provided: (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.

3n-2)/n^3My 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.

Appreciate any help.