ITPEC IP April 2026 Question 3
When a chairperson and a secretary are selected from three (3) candidates with a method that allows dual roles, there are three (3) ways for the selection of a chairperson and three (3) ways for the selection of a secretary, so in total, there are nine (9) combinations for the selection of a chairperson and a secretary. When a chairperson and a secretary are selected from five (5) candidates with a method that allows dual roles, how many combinations for the selection are there?
Explanation
25 — each role is filled independently from all five candidates, so multiply 5 × 5.
Because one person may hold both roles, choosing the chairperson does not remove anyone from the choice of secretary. The example in the question works the same way: 3 × 3 = 9. With five candidates there are 5 choices for chairperson and still 5 for secretary: 5 × 5 = 25.
Answer (d)
Why not others:
- •(a) 5 — counts the choices for one role only
- •(b) 10 — is 5 + 5, adding the choices instead of multiplying, and also the number of unordered pairs of two different people, which ignores both the distinct roles and dual roles
- •(c) 20 — is 5 × 4, the count when the secretary must be someone other than the chairperson, which the question does not require
Key rule: Independent choices multiply; when the same item may be chosen again (repetition allowed), each position has the full n choices, giving nk arrangements.
Basic Theory · AI-written, may contain errors
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