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09 Jun 2026, 06:36
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
44% (02:22) correct
56%
(01:46)
wrong
based on 9
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Armstead Productions conducted two rounds of interviews for 5 candidates. In each round, each candidate was assigned at random to one of the company’s 5 managers so that each manager interviewed exactly one candidate.
What is the probability that exactly 3 of the 5 managers interviewed more than one of the 5 candidates ?
(A) \(\frac{1}{12}\)
(B) \(\frac{1}{10}\)
(C) \(\frac{1}{8}\)
(D) \(\frac{2}{15}\)
(E) \(\frac{1}{6}\)
What is the probability that exactly 3 of the 5 managers interviewed more than one of the 5 candidates ?
(A) \(\frac{1}{12}\)
(B) \(\frac{1}{10}\)
(C) \(\frac{1}{8}\)
(D) \(\frac{2}{15}\)
(E) \(\frac{1}{6}\)
ankushsambare
Joined: 13 Jun 2022
Last visit: 10 Jun 2026
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Schools: ISB '27 Wharton '26 INSEAD Aug '26
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09 Jun 2026, 20:01
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Think of the first round as fixed.
In the second round, each of the 5 candidates is again assigned to the 5 managers, one candidate per manager. Thus the second round corresponds to a permutation of the 5 candidates.
A manager interviews the same candidate in both rounds iff that manager is a fixed point of the permutation.
The question asks for the probability that exactly 3 managers interviewed more than one candidate.
Since each manager interviews exactly one candidate in each round:
Exactly 3 managers interviewed more than one candidate
⇒ exactly 2 managers interviewed the same candidate both rounds
⇒ the permutation has exactly 2 fixed points.
Total permutations of 5 candidates:
5! = 120
Count permutations with exactly 2 fixed points:
10 × 2 = 20
Probability:
20 / 120 = 1/6
Answer: E. 1/6.
In the second round, each of the 5 candidates is again assigned to the 5 managers, one candidate per manager. Thus the second round corresponds to a permutation of the 5 candidates.
A manager interviews the same candidate in both rounds iff that manager is a fixed point of the permutation.
The question asks for the probability that exactly 3 managers interviewed more than one candidate.
Since each manager interviews exactly one candidate in each round:
- If a manager gets the same candidate both times, they interviewed only 1 distinct candidate.
- If a manager gets a different candidate in round 2, they interviewed 2 distinct candidates.
Exactly 3 managers interviewed more than one candidate
⇒ exactly 2 managers interviewed the same candidate both rounds
⇒ the permutation has exactly 2 fixed points.
Total permutations of 5 candidates:
5! = 120
Count permutations with exactly 2 fixed points:
- Choose the 2 fixed points:
C(5,2) = 10 - The remaining 3 candidates must have no fixed points (a derangement of 3).
Number of derangements of 3:
!3 = 2
10 × 2 = 20
Probability:
20 / 120 = 1/6
Answer: E. 1/6.
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