Floor-plan sign-off: five officers, five floors
Five officers of a cooperative bank's zonal office occupy five floors numbered 1 (ground) to 5, one officer per floor. The officers are P, Q, R, S and T, and each heads exactly one department: Credit, Audit, Forex, HR and Legal. The floor plan obeys these clues: (i) P is on floor 3. (ii) S is on an even-numbered floor and heads HR. (iii) The Forex head sits immediately above the Legal head. (iv) Q heads Credit and sits immediately below T. (v) R does not head Legal. The administrative officer wants a seating-cum-department chart before the 4 p.m. audit briefing. A colleague has already written P-Legal, T-Forex and S-HR on the whiteboard and wants your sign-off; no working is attached, and the chart must survive a challenge if any single clue is questioned at the briefing. You are also told the vendor can add a sixth floor later, so the logic, not just the answer, will be reused.
1. Prepare the complete floor-and-department chart, showing how each alternative is eliminated.
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Step 1 - Place Q and T. Clue (iv) allows (Q,T) = (1,2), (2,3), (3,4), (4,5). Clue (i) puts P on floor 3, and P is a different person from both, so (2,3) fails (T = 3) and (3,4) fails (Q = 3). Two branches survive.
Step 2 - Branch I: Q = 1 (Credit), T = 2. S must be even and cannot be 2, so S = 4 (HR). Floors 3 and 5 go to P and R, and P is fixed at 3, so R = 5. Departments left - Audit, Forex, Legal - must fill floors 2, 3, 5. Clue (iii) needs an adjacent pair, and only (2,3) qualifies: T = Forex, P = Legal, leaving R = 5 Audit. Clue (v) is satisfied.
Step 3 - Branch II: Q = 4 (Credit), T = 5. S must be even and cannot be 4, so S = 2 (HR). P = 3, hence R = 1. The three remaining departments would have to sit on floors 1, 3 and 5 - no two of which are adjacent - so clue (iii) cannot be met. Branch II dies.
Chart: Floor 1 Q (Credit); Floor 2 T (Forex); Floor 3 P (Legal); Floor 4 S (HR); Floor 5 R (Audit). The colleague's P-Legal, T-Forex and S-HR entries are correct, but they are outputs of the reasoning, not assumptions that can be signed off on their own.
2. Which single clue is dispensable because the remaining clues already force a unique chart? Prove it, and state why the other clues cannot be dropped.
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Clue (v), 'R does not head Legal', is redundant.
With (i)-(iv) and (ii) kept, Branch I fixes every floor, and the only adjacent pair available for Forex/Legal is (2,3). That forces T = Forex and P = Legal, so R on floor 5 is left with Audit automatically - clue (v) never has to be used. Branch II is rejected on adjacency alone, before any department is assigned to R. Exactly one chart survives, so (v) adds nothing.
Why the others are load-bearing. Drop (iii) and the count explodes: in Branch I the three departments Audit/Forex/Legal can occupy floors 2, 3, 5 in 3! = 6 ways, of which 2 put Legal on floor 5 (barred by (v)), leaving 4; in Branch II floors 1, 3, 5 give 6 ways, of which 2 put Legal on floor 1 (barred by (v)), leaving 4. That is 8 valid charts, not 1. Drop (ii) and S could sit on floor 2 in Branch I or floor 4 in Branch II, reopening Branch II. Drop (iv) and Q/T placement loses its anchor entirely. So the examinable point is that a clue can be redundant even when it looks substantive - redundancy is a property of the whole set, not of the sentence.
3. Your colleague solved the puzzle by assuming S is on floor 4 rather than on an even-numbered floor. Is that shortcut safe to present at the briefing?
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It produces the right chart here, but it is not a valid method and should not be presented as one.
Assuming S = 4 kills Branch II at once (in Branch II S = 2), so the colleague reaches the correct final chart by deleting a live branch rather than testing it. The shortcut only 'works' because the branch it discards happens to be the branch that later fails on clue (iii). Change one input - say clue (iv) allowed Q = 4, T = 5 with S forced to 2 - and the assumption would silently discard the only valid solution. On a floor-plan question the discarded branch is not visible in the answer, so the error would go undetected.
What to present: the two-branch elimination. Branch II is the evidence that the clue set is consistent and has a unique solution; without it, the briefing cannot answer the obvious challenge 'how do you know there is not another arrangement?' Keep the derived value S = 4 as a conclusion with the line 'Branch II fails clue (iii)' attached, and note for the future sixth-floor extension that S would then have floor 6 as an additional even option, so the branch count must be redone rather than patched.