The Branch Immersion Roster That Was Declared 'Fixed' Too Early
The HR department of a mid-sized private bank must schedule five campus-hired probationary officers - A, B, C, D and E - for a one-day branch immersion between Monday and Friday, exactly one candidate per day. The rules are: (i) A must go on Wednesday; (ii) B must be placed on the day immediately before E; (iii) C cannot be placed on Monday; (iv) D cannot be placed on Friday. A junior coordinator prepares a chart and declares the schedule "fully fixed". The HR head is not convinced and orders an audit before the appointment letters are dispatched, because any rescheduling after dispatch means reissuing five letters and rebooking travel worth about Rs 1.4 lakh. The coordinator also proposes adding one more rule - "E cannot be placed on Monday" - in the belief that this will settle the matter. No two candidates may share a day, and all five must be accommodated within the five working days.
1. How many valid schedules exist? Enumerate them and show the deduction rather than guessing.
Read the worked answer
Step 1 - Fix the certain slot. Rule (i) puts A on Wednesday (day 3).
Step 2 - Place the B–E block. B must be immediately before E, so the block occupies two consecutive days. Because day 3 is taken by A, the only possible blocks are (day 1, day 2) or (day 4, day 5).
Step 3 - Fill the rest in each case.
- Case 1: B = Mon, E = Tue. Remaining slots are Thu and Fri for C and D. Rule (iv) bars D from Friday, so D = Thu and C = Fri. Check rule (iii): C is not on Monday - satisfied. Schedule 1: B, E, A, D, C.
- Case 2: B = Thu, E = Fri. Remaining slots are Mon and Tue for C and D. Rule (iii) bars C from Monday, so C = Tue and D = Mon. Check rule (iv): D is not on Friday - satisfied. Schedule 2: D, C, A, B, E.
Conclusion: exactly two valid schedules exist. The coordinator's chart is one of them, but it is not the only permissible one, so the claim "fully fixed" is false.
2. Does the coordinator's proposed extra rule, "E cannot be placed on Monday", make the schedule unique? Justify.
Read the worked answer
Test the rule against both surviving schedules.
- Schedule 1: E is on Tuesday.
- Schedule 2: E is on Friday.
In neither schedule is E on Monday, so the proposed rule is already satisfied by both schedules. A clue that is true in every permissible arrangement removes no arrangement; it is redundant, not decisive. The count therefore remains two, and the roster stays under-determined.
What would actually work? Any clue that splits the two cases, i.e. one that fixes the B–E block. For example, "D is on Monday" forces Schedule 2; "C is on Friday" forces Schedule 1; "B is on Thursday" forces Schedule 2. The design principle is worth remembering for every arrangement puzzle: before accepting an added clue, test it against the full solution set - a clue is useful only if at least one valid arrangement violates it.
3. The HR head has four candidate clues: (i) E is not on Thursday; (ii) C is on Friday; (iii) D is on Monday; (iv) B is on Thursday. Which of these will NOT make the schedule unique, and why?
Read the worked answer
Evaluate each clue against Schedule 1 (B, E, A, D, C) and Schedule 2 (D, C, A, B, E).
- (i) E is not on Thursday. E is on Tuesday in Schedule 1 and on Friday in Schedule 2 - both satisfy the clue. Two arrangements survive, so the roster is still not unique. This is the answer.
- (ii) C is on Friday. True only in Schedule 1, which is therefore forced. Unique.
- (iii) D is on Monday. True only in Schedule 2, which is forced. Unique.
- (iv) B is on Thursday. True only in Schedule 2, which is forced. Unique.
Rule to carry forward: a clue resolves an arrangement only when it is true of exactly one member of the solution set. Clues that are true of all solutions (like (i), and like the coordinator's "E not on Monday") add words but no information; clues true of none signal that an earlier deduction was wrong and the premises must be re-checked.