The Six-Chair Roster
The newly opened Chikhli branch has six counter chairs in a single row, numbered 1 to 6 from left to right, all facing the customer side. Six officers - A, B, C, D, E and F - must be allotted one chair each for the morning shift. The branch head circulates five conditions: (i) A sits third to the left of D; (ii) E sits immediately to the right of A; (iii) C sits at one of the two ends; (iv) B sits next to D; (v) B does not sit next to E. A trainee submits a single seating plan. The branch head refuses to sign the roster, saying the plan 'cannot be the only one', and will not approve it until the trainee either proves the arrangement is unique or produces every valid plan. The roster must be printed in two hours. In these conditions, 'third to the left' means exactly three chairs away in the leftward direction, and 'next to' means immediately adjacent.
1. How many seating arrangements satisfy all five conditions? List each valid arrangement, showing how the A-D pair is fixed before the remaining officers are placed.
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Step 1 - the A-D pair is the anchor. Condition (i) allows only three placements: A1-D4, A2-D5 and A3-D6. Condition (ii) then fixes E at 2, 3 and 4 respectively.
Step 2 - test each case against (iii), (iv), (v).
Case A1-D4-E2: C must take an end and chair 1 is A's, so C = 6. Chairs 3 and 5 are free for B and F. B must be next to D(4), so B is 3 or 5. But B = 3 is next to E(2), which (v) forbids. So B = 5, F = 3. Arrangement: 1-A, 2-E, 3-F, 4-D, 5-B, 6-C.
Case A2-D5-E3: C is 1 or 6. If C = 1, chairs 4 and 6 are free; B must be next to D(5), so B = 4 or 6; B = 4 sits next to E(3) and is rejected, leaving B = 6, F = 4 - valid: 1-C, 2-A, 3-E, 4-F, 5-D, 6-B. If C = 6, B can only be 4, again next to E(3) - rejected.
Case A3-D6-E4: C can only be 1; chairs 2 and 5 remain; B must be 5 to sit next to D(6), but 5 is next to E(4) - rejected.
Answer: exactly two plans, so the trainee's single plan was not proved unique.
2. State one additional condition that would make the roster unique, and prove that it works by testing it against both arrangements. Also show one plausible extra condition that would fail.
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The extra condition must separate the two surviving plans, which differ in three places: plan 1 has A at 1, C at 6, B at 5; plan 2 has A at 2, C at 1, B at 6. A useful clue must be false in one plan and true in the other.
Working condition: 'C sits to the left of A.' In plan 1, C is at 6 and A at 1, so C is to the right of A - the condition kills plan 1. In plan 2, C is at 1 and A at 2, so C is to the left of A - plan 2 survives. Only one plan remains, so this condition makes the answer unique.
Condition that fails: 'F sits to the left of D.' In plan 1, F is at 3 and D at 4 - true. In plan 2, F is at 4 and D at 5 - also true. The clue is satisfied by both plans and changes nothing, so it is wasted.
Method note: do not accept an extra clue simply because it 'sounds helpful'. Before recommending it, run it against every arrangement already found. A condition that both plans satisfy, or that both violate, adds no information about this pair.
3. The trainee claims condition (v) is unnecessary and could be dropped. Test the claim.
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Claim is false - (v) is load-bearing.
Remove (v) and re-run only the A1-D4-E2 case, which is where (v) did the work. With A at 1, E at 2, D at 4 and C at 6, chairs 3 and 5 are free for B and F. Condition (iv) only requires B to be next to D at 4, and both 3 and 5 qualify. That yields two plans instead of one:
- 1-A, 2-E, 3-B, 4-D, 5-F, 6-C
- 1-A, 2-E, 3-F, 4-D, 5-B, 6-C (the one found earlier)
The other two cases are unaffected by (v), so the puzzle would move from two solutions to three.
General rule to apply here: a clue is redundant only if deleting it leaves the solution set unchanged. The test is to re-solve without it and compare the count, not to judge whether the clue 'looks obvious'. Because deleting (v) adds a solution, the trainee's claim must be rejected and (v) must stay in the circulated puzzle.