The Five-Day Interview Puzzle with One Bad Clue
A bank's HR team must schedule five candidates-A, B, C, D and E-for one interview each, Monday to Friday. Each candidate must occupy a different day, and no candidate can be moved to a weekend. Two coordinators supplied five clues: (1) A is interviewed on Wednesday. (2) B is interviewed immediately before D. (3) C is not interviewed on Monday or Friday. (4) E is interviewed after A. (5) D is interviewed immediately before C. The panel knows that exactly one of these five clues is wrong, but not which. The Monday slot is already printed on an offer letter, so the team cannot leave any day blank. A junior assistant claims the puzzle has no solution and wants to re-interview all candidates. The HR head asks you to test the clues rather than discard the schedule.
1. If clue (5) is the erroneous one, build the schedule and test uniqueness.
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With (5) removed, the live clues are (1)–(4). A is fixed on Wednesday. B must be immediately before D, so the B–D pair can only be Monday–Tuesday or Thursday–Friday. If it were Thursday–Friday, C and E would be left for Monday–Tuesday; E after A would then be impossible because both remaining days precede Wednesday. So B–D is Monday–Tuesday. That leaves Thursday and Friday for C and E. Clue (3) rules out Friday for C, so C is Thursday and E is Friday. Schedule: Mon B, Tue D, Wed A, Thu C, Fri E. It satisfies all four live clues and is the only arrangement under them, so removing (5) gives a unique schedule.
2. If clue (2) is erroneous, is the assistant's 'no solution' claim correct? Show the possibilities.
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Remove (2). Live clues: A on Wednesday; C not Mon/Fri; E after A; D immediately before C. The D–C pair can be Mon–Tue or Thu–Fri. If Thu–Fri, then B and E occupy Mon–Tue, and E cannot be after A, so impossible. Thus D is Monday and C Tuesday. Thursday and Friday remain for B and E. E after A means E can be Thursday or Friday, giving two schedules: (i) Mon D, Tue C, Wed A, Thu E, Fri B; (ii) Mon D, Tue C, Wed A, Thu B, Fri E. Both satisfy the live clues. So the assistant is wrong that there is no solution; the real problem is non-uniqueness, not impossibility.
3. Which single clue should the HR head treat as the erroneous one if a unique schedule is required? Justify briefly by elimination.
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Treat clue (5) as erroneous. Removing (5) gives the unique schedule in Task 1. Removing (2) leaves two schedules. Removing (1) leaves multiple valid schedules, e.g. B,D,C,A,E and A,B,D,C,E. Removing (3) or (4) does not rescue uniqueness: with (5) live, B–D and D–C force B, D, C into three consecutive days, and A is fixed on Wednesday; every three-day block in a five-day week contains Wednesday, so A collides with one of them. Those removals yield no solution, not a unique one. Only removal of (5) yields exactly one valid schedule.