Shelf Order in the Record Room: Testing Five Conditions Against One Claim
The record room of a Delhi Police station is stacking five bound registers - A, B, C, D and E - on five numbered shelves of a steel almirah. Shelf 1 is the bottom shelf and shelf 5 is the top; exactly one register goes on each shelf. The storekeeper's handover note records five conditions: (i) A is immediately above C; (ii) B is immediately above E; (iii) D is not on shelf 1; (iv) D is somewhere above A; and (v) D is immediately above B. A newly posted constable has arranged the registers as C on 1, A on 2, B on 3, D on 4 and E on 5, and claims the arrangement satisfies the note. The inspector, before signing the handover, wants each condition tested separately and the arrangement that actually fits all five conditions established, because a wrong shelf order will slow down the tracing of old registers for the whole of the next month. The almirah cannot be expanded, and no two registers may share a shelf.
1. Test the constable's arrangement against each of the five conditions and then derive the arrangement that satisfies all of them.
Read the worked answer
Test the constable (1 C, 2 A, 3 B, 4 D, 5 E):
- (i) A on 2 is immediately above C on 1 - satisfied.
- (ii) B on 3 must be immediately above E - E is on 5, so B is below E. Violated. One violation is enough to reject the arrangement.
- (iii) D on 4 is not on shelf 1 - satisfied.
- (iv) D on 4 is above A on 2 - satisfied.
- (v) D on 4 is immediately above B on 3 - satisfied.
So the claim fails only on (ii), which is exactly the kind of single-condition miss a handover check exists to catch.
Derivation: (ii) and (v) together force the three-shelf block D, B, E in that descending order, so the block must occupy shelves (3,2,1), (4,3,2) or (5,4,3).
- Block on (3,2,1): shelves 4 and 5 remain for A and C; A immediately above C forces A = 5, C = 4. But then D on 3 is below A, violating (iv).
- Block on (4,3,2): shelves 1 and 5 remain for A and C, which are not consecutive, so (i) cannot hold.
- Block on (5,4,3): shelves 1 and 2 remain, giving A = 2, C = 1. Check (iv): D on 5 is above A on 2 - satisfied. Check (iii): D is not on shelf 1 - satisfied.
Unique valid order: 1 C, 2 A, 3 E, 4 B, 5 D.
2. The inspector asks a weaker question: if condition (v) is replaced by 'D is somewhere above B' (no longer immediately), how many arrangements satisfy the remaining conditions?
Read the worked answer
Keep (i) A immediately above C, (ii) B immediately above E, (iii) D not on 1, (iv) D above A, and replace (v) with 'D above B'. Enumerate by the position of the pair A–C, since that pair fixes two shelves at once.
- A = 2, C = 1. Remaining shelves 3, 4, 5. The pair B–E can sit at (4,3) or (5,4).
- B = 4, E = 3 → D = 5. D above A (2) and above B (4): both true. Valid: 1 C, 2 A, 3 E, 4 B, 5 D.
- B = 5, E = 4 → D = 3. D above A yes, but D on 3 is below B on 5. Rejected.
- A = 3, C = 2. Remaining 1, 4, 5. The pair B–E can only sit at (5,4), leaving D = 1, which violates (iii) and also puts D below A. Rejected.
- A = 4, C = 3. Remaining 1, 2, 5. The pair B–E can only sit at (2,1), leaving D = 5. D above A (4) and above B (2): both true. Valid: 1 E, 2 B, 3 C, 4 A, 5 D.
- A = 5, C = 4. Then D cannot be above A, so (iv) fails.
Count = 2 arrangements, not 1. The original condition (v) was doing real work: it removed the second solution.
3. The clerk now argues that if condition (v) is dropped entirely - so D has no stated relation to B - there would be 'more than four' valid arrangements. Test this claim and state the exact number.
Read the worked answer
Conditions now: A immediately above C; B immediately above E; D not on shelf 1; D above A.
Enumerate by the A–C pair again:
- A = 2, C = 1. Remaining 3, 4, 5. B–E at (4,3) → D = 5, and D is above A: valid. B–E at (5,4) → D = 3, D above A: valid. 2 arrangements.
- A = 3, C = 2. Remaining 1, 4, 5. B–E at (5,4) → D = 1, which breaks (iii) and is below A. B–E at (2,1) is impossible because shelf 2 is taken by A. 0 arrangements.
- A = 4, C = 3. Remaining 1, 2, 5. B–E at (2,1) → D = 5, above A: valid. B–E at (5,4) is impossible because shelf 4 is taken by A. 1 arrangement.
- A = 5, C = 4. D cannot be above A. 0 arrangements.
Total = 2 + 0 + 1 + 0 = 3 arrangements. The claim of 'more than four' is false, and it is false in the safe direction: removing a constraint can only increase the count, so the clerk's error is arithmetic in the enumeration, not a misunderstanding of the conditions. The practical lesson for the handover note is that three of the five original conditions were load-bearing; drop any one of them and the shelf order stops being reproducible by the next shift.