Five-Locker Allocation in the Narcotics Cupboard
You are the store in-charge of a district hospital pharmacy. The controlled-drugs cupboard has five lockers, numbered 1 to 5 counting from the door, and each locker is assigned for the month to exactly one of five staff: A, B, C, D and E. The standing instructions impose these conditions. A must have locker 1, because she holds the key to the main grill. C is on provisional registration and may not stand next to the entrance, so she cannot take locker 1 or locker 5. E, who prepares the night audit, must have locker 5. B and D must be immediately next to each other, so that two-person verification of stock is possible. B must be closer to the door than D. C and E must not be adjacent, because their record books are combined at the close of day. The cupboard also holds narcotic drugs, all of which are kept in the double-locked inner cabinet; some Schedule H1 drugs are also kept in that inner cabinet for security.
1. Determine the complete locker allocation and show how the conditions eliminate the alternatives.
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Fixed points first. A = 1 and E = 5 are given directly. Lockers 2, 3 and 4 must therefore go to B, C and D.
Apply the C condition. C cannot be at 1 or 5, which is already satisfied, and C must not be adjacent to E at 5 - so C cannot be at 4. C is therefore at 2 or 3.
Test C = 3. Then B and D occupy 2 and 4, which are not adjacent. The adjacency condition fails. C ≠ 3.
Test C = 2. Then B and D occupy 3 and 4, which are adjacent. B must be nearer the door, so B = 3 and D = 4. Check C and E: lockers 2 and 5 are not adjacent, so that condition also holds.
Result: A–1, C–2, B–3, D–4, E–5. Every stated condition is satisfied and no other arrangement survives, so the allocation is unique - which is exactly what you want in a controlled-drugs store, because a unique allocation makes an unexplained key or an unaccounted entry immediately visible at audit.
2. Suppose the condition "B must be closer to the door than D" is withdrawn. How many valid allocations now exist, and what are they?
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Keep A = 1 and E = 5 fixed, so B, C and D occupy 2, 3 and 4, with C not adjacent to E and therefore C ≠ 4. C is at 2 or 3.
Case C = 2. B and D take 3 and 4, which are adjacent. Two orders are possible: B–3, D–4 and D–3, B–4. That is 2 allocations.
Case C = 3. B and D take 2 and 4, which are not adjacent, so this case fails entirely.
Total: 2 allocations, namely A–1, C–2, B–3, D–4, E–5 and A–1, C–2, D–3, B–4, E–5.
The reasoning point is that a directional clue such as "closer to the door" removes exactly one degree of freedom. Remove it and the number of solutions doubles rather than staying the same, because adjacency is symmetric - it is satisfied by either ordering of the pair. In a real store this matters: a roster that permits two equally valid allocations cannot be used to prove who was responsible for a particular locker on a particular day unless the written allotment order is signed.
3. Using the cupboard facts - all narcotic drugs are in the double-locked cabinet, and some Schedule H1 drugs are also in that cabinet - decide which of these conclusions follow: (i) some Schedule H1 drugs are kept in the double-locked cabinet; (ii) some narcotic drugs are Schedule H1 drugs; (iii) some Schedule H1 drugs in the cupboard are not narcotic drugs.
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Conclusion (i) follows. The premise is "some Schedule H1 drugs are in the double-locked cabinet". A particular affirmative statement converts simply: if some H1 drugs are in the cabinet, then some items in the cabinet are H1 drugs. This is the same fact read in the other direction, not an inference beyond the data.
Conclusion (ii) does not follow. From "all narcotics are in the cabinet" and "some H1 drugs are in the cabinet" nothing can be said about the overlap between narcotics and H1 drugs. The H1 drugs in the cabinet may be entirely separate from the narcotics. At most one can draw the possibility, never the certainty.
Conclusion (iii) does not follow. It asserts that at least one H1 drug in the cupboard is outside the narcotic set. The premises are consistent with every H1 drug in the cabinet being a narcotic as well, so the conclusion is only a possibility.
The standard error in this type of question is to treat "some" as "some but not all". In logic, "some" means "at least one" and permits overlap with the whole; that single point decides all three parts.