Five-officer seating puzzle: is the junior's arrangement unique?
Five officers-A, B, C, D and E-are to be seated in a single row facing north for a photograph. The seating committee has issued these clues: (i) A sits exactly in the middle. (ii) B sits immediately to the left of D. (iii) C does not sit at either end. (iv) E does not sit adjacent to A. A junior assistant concludes that the only arrangement is E–C–A–B–D (left to right) and prepares the final seating chart. A senior officer suspects there may be a second valid arrangement. The committee cannot release a chart with two possibilities because the photograph protocol assigns a fixed position to each officer by name. The chair asks for a written note before 4 p.m. The note must state whether the clues determine a unique arrangement, show all valid arrangements, and if not unique, propose the smallest additional condition that would make it unique without contradicting the existing clues. No new clue may name a position directly if a relational clue can do the job.
1. Test the junior's conclusion by enumerating all arrangements that satisfy clues (i)–(iv). Is the chart unique?
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Fix positions 1–5 from left to right. Clue (i) gives A = 3. Clue (iii) says C is not at an end, and position 3 is occupied, so C = 2 or C = 4.
Case 1: C = 2. Remaining positions are {1, 4, 5} for B, D and E. Clue (ii) requires B immediately left of D, so the pair must occupy consecutive positions. Only (4, 5) is available: B = 4, D = 5, leaving E = 1. Check (iv): E at 1 and A at 3 are not adjacent. Valid: E C A B D.
Case 2: C = 4. Remaining positions are {1, 2, 5}. The only consecutive pair available is (1, 2), so B = 1, D = 2, leaving E = 5. Check (iv): E at 5 and A at 3 are not adjacent. Valid: B D A C E.
There are two valid arrangements. The junior's chart is one of them, not the unique solution. The junior probably fixed C = 2 without testing C = 4.
2. Propose the smallest additional condition that makes the solution unique without naming a position directly. Show that it eliminates exactly one arrangement and does not contradict the clues.
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Add: 'B sits to the right of C.'
- In E C A B D, B is at 4 and C at 2, so B is to the right of C - this arrangement is kept.
- In B D A C E, B is at 1 and C at 4, so B is to the left of C - this arrangement is eliminated.
The remaining arrangement is unique: E C A B D. This is one relational clue, so it is minimal in the sense that a single added condition separates the two cases; adding two clues would be unnecessary. It does not name a position. Another valid one-clue alternative is 'D sits to the right of A': it keeps Case 1 (D = 5, A = 3) and eliminates Case 2 (D = 2, A = 3). Whichever is chosen, verify that both original cases satisfy the old clues and that the new clue is not already implied by them.
3. If the puzzle was already printed in a mock test with the instruction 'unique solution', what is the correct course of action?
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Do not defend the ambiguity. The defensible options are:
- Issue a corrigendum adding the new clue and re-release the answer key with E C A B D.
- If the test is already over, mark both arrangements as correct and award full marks to candidates who chose either, because both satisfy the printed clues.
- For future papers, run a uniqueness check by enumerating all permutations or by using a solver before printing.
For the photograph itself, do not let an ambiguous puzzle decide the seating. Use an independent tie-breaker, such as seniority, and record it. The key is transparency: a false claim of uniqueness damages the exam's credibility more than a corrected key does.