The schedule that was signed off as 'fully determined'
A regional office must fix the schedule for six insurance agents - A, B, C, D, E and F - who will each spend one full working day, Monday to Saturday, at a branch. Exactly one agent goes each day, and each agent goes on exactly one day. The branches themselves do not enter the logic; only the days matter. The HR executive has circulated five conditions: (i) C goes on Wednesday; (ii) A goes on the day immediately before B; (iii) D does not go on Monday or Saturday; (iv) E does not go on a day adjacent to C's day; (v) F goes after C. The covering note says the schedule is 'fully determined' and asks you to sign it off before 5 p.m. Two agents have already booked non-refundable travel, one for Monday and one for Tuesday, and any change costs ₹4,500 per ticket. The executive proposes that, to break any residual tie, those two agents should simply be placed on Monday and Tuesday.
1. Is the schedule actually fully determined? Establish the number of valid schedules and show the case split.
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The note is wrong. Because C is fixed at Wednesday (slot 3), the consecutive pair A–B can sit only at (1,2), (4,5) or (5,6) - never at (2,3) or (3,4), which contain Wednesday.
Pair at (1,2): A1, B2. D must be 4 or 5 by condition (iii).
- D = 4 → two arrangements (E5 F6, and E6 F5)
- D = 5 → E must avoid slots 2 and 4, so E = 6 and F = 4 → one arrangement Three here.
Pair at (4,5): A4, B5. Free slots 1, 2, 6. D cannot be 1 or 6, so D = 2. E cannot be 2 or 4, so E = 1. F must be after C, so F = 6. One arrangement.
Pair at (5,6): A5, B6. Free slots 1, 2, 4. D = 2 or 4; E cannot be 2 or 4, so E = 1; F after C forces F = 4, hence D = 2. One arrangement.
Total = 5 schedules:
- E, D, C, F, A, B
- E, D, C, A, B, F
- A, B, C, D, E, F
- A, B, C, D, F, E
- A, B, C, F, D, E
Signing the note as it stands would lock in an arrangement nobody has verified.
2. Which single added condition would make the schedule unique? Test 'F goes on Thursday', 'D goes on Friday' and 'A goes on Monday' before concluding.
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Test each candidate against the five schedules above.
F on Thursday (slot 4): appears in schedules 1 and 5 - two solutions, still ambiguous.
D on Friday (slot 5): appears only in schedule 5 (A, B, C, F, D, E). Unique.
A on Monday: appears in schedules 3, 4 and 5 - three solutions, ambiguous.
So 'D goes on Friday' works, and it works because it kills the entire (4,5) and (5,6) branches of the case split in one stroke; the other two candidates only trim part of one branch.
A good answer does not stop at one success: the point of the test is that uniqueness has to be proved by exhaustion, not by finding one clue that 'feels' decisive. Any candidate condition that only eliminates one schedule is useless, because four remain.
3. Is the executive's tie-break proposal logically safe? State what the five conditions already force about Monday, Tuesday and Wednesday, and what that implies for the two travellers.
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The five conditions already force a great deal, and the proposal must be checked against it rather than against convenience.
- Wednesday is C in every solution.
- Monday can only be A or E.
- Tuesday can only be B or D.
Therefore the proposal is safe only if the two travellers are, in some order, one of {A, E} and one of {B, D}. If the travellers are, say, C and F, the proposal is not merely untidy - it is impossible, and the ₹4,500 per-ticket cost would have been incurred for nothing.
Even when the travellers do fit, the proposal does not remove the ambiguity: A on Monday is compatible with three of the five schedules. A tie-break must be a logical condition that eliminates schedules, not an administrative preference dressed up as one.
The defensible course is to tell the executive the true status - five valid schedules, Wednesday fixed, Monday restricted to {A, E}, Tuesday restricted to {B, D} - and only then spend money on travel.