問72
頻出度★★☆
3人の候補者の中から兼任も許す方法で委員長と書記を1名ずつ選ぶ場合, 3人の中から委員長1名の選び方が3通りで, 3人の中から書記1名の選び方が3通りであるので, 委員長と書記の選び方は全部で9通りある。5人の候補者の中から兼任も許す方法で委員長と書記を1名ずつ選ぶ場合, 選び方は何通りあるか。
令和元年度 秋期 ITパスポート試験 問72
テクノロジ系基礎理論 › 応用数学
This question's explanation and your answers so far.
解説
問 72
Incorrectア
5 is merely the number of candidates, or the number of ways to choose just one person; it does not count the combinations for selecting the two roles of chairperson and secretary, so it is incorrect.
Incorrectイ
10 corresponds to the number of combinations for choosing 2 people out of 5 without distinguishing the roles (5×4÷2=10). Since the chairperson and secretary are distinct roles and concurrent appointment is allowed, this counting method does not apply, so it is incorrect.
Incorrectウ
20 is 5×4=20, the number of ways when the chairperson and secretary must be different people (concurrent appointment not allowed). Since this question allows concurrent appointment, the 5 cases where the same person holds both posts must also be included; 20 is too few, so it is incorrect.
- エ
正解 Because concurrent appointment is allowed, the ways to choose the chairperson and the ways to choose the secretary are independent, with 5 ways each. By the multiplication rule, 5 ways for chairperson × 5 ways for secretary = 25 ways. This is the same reasoning as the 3-person example in the question (3×3=9 ways); only the number of candidates has increased to 5. Therefore, the correct answer is エ (25).
ここがポイント
- Multiplication rule: when independent choices are made in sequence, the total number of cases equals the product of the number of ways at each stage (5 ways for chairperson × 5 ways for secretary = 25 ways).
- Concurrent appointment allowed = the same person may hold both roles, so each role can be chosen independently from 5 options. If concurrent appointment is not allowed, it becomes 5×4=20 ways; distinguishing this point is important.
用語