Invigilator Seating Plan for a Five-Desk Row
An examination centre has to seat five candidates-A, B, C, D, and E-at desks numbered 1 to 5 in a single row from left to right. The centre has one invigilator for the row and no spare desk. The centre coordinator has fixed the following rules: A cannot sit at either end desk. B must sit immediately to the left of C. D must sit at desk 4. E must not sit on a desk adjacent to A. C cannot sit at desk 5. The coordinator also wants to know whether the plan is unique and what would change if D were shifted to desk 5. The five candidates have no other restrictions, and all desks are single-seat. The centre must finalise the plan before the answer sheets are distributed.
1. Determine the complete seating arrangement from desks 1 to 5. Show each deduction.
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D is fixed at desk 4. B must be immediately left of C, so the possible B–C pairs are (1,2), (2,3), (3,4), and (4,5). Pairs (3,4) and (4,5) are impossible because D already occupies desk 4. If B–C were (2,3), then A would have to take desk 1 or 5, since desks 2, 3, and 4 are occupied-but A cannot sit at either end. So (2,3) fails. Therefore B–C = (1,2): B at 1, C at 2. Remaining desks are 3, 4, 5 for A, D, E. D = 4. A cannot be at 5 (end), so A = 3. E = 5. Check E not adjacent to A: desks 5 and 3 are not adjacent. Arrangement: B, C, A, D, E.
2. If D is moved to desk 5, is there still a valid arrangement? Give one or explain why none exists.
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With D = 5, test the B–C pairs. (1,2): B=1, C=2; remaining desks 3,4,5 for A,D,E; D=5; A cannot be 1 or5, so A=3 or4. If A=3, E=4, and E is adjacent to A, violating the rule. If A=4, E=3, again adjacent. So (1,2) fails. (2,3): B=2, C=3; remaining 1,4,5; D=5; A cannot be 1 or5, so A=4; E=1. E at 1 is not adjacent to A at 4. This works: E, B, C, A, D. (3,4): B=3, C=4; remaining 1,2,5; D=5; A cannot be 1 or5, so A=2; E=1, but E is adjacent to A, so fails. Thus exactly one valid arrangement exists: E, B, C, A, D. Moving D to desk 5 does not make the plan impossible, but it changes the unique solution.
3. If the rule that A cannot sit at either end is removed, list all valid arrangements under the other rules.
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Keep D=4, B immediately left of C, C not at 5, E not adjacent to A. The B–C pairs are still (1,2) or (2,3) because (3,4) and (4,5) collide with D. For (1,2): B=1, C=2; remaining 3,4,5; D=4; A can be 3 or5. If A=3, E=5; E not adjacent to A (5 vs 3) works: B, C, A, D, E. If A=5, E=3; E not adjacent to A (3 vs 5) works: B, C, E, D, A. For (2,3): B=2, C=3; remaining 1,4,5; D=4; A can be 1 or5. If A=1, E=5; not adjacent: A, B, C, D, E. If A=5, E=1; not adjacent: E, B, C, D, A. So four arrangements: B C A D E; B C E D A; A B C D E; E B C D A.