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Row of Six Counters: Is the Posting Unique?

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Row of Six Counters: Is the Posting Unique?

Six probationary officers - A, B, C, D, E and F - are to be posted at the six teller counters of a newly opened branch. The counters stand in one row facing customers and are numbered 1 to 6 from the left. The branch head has fixed five conditions. A must sit exactly three counters to the left of D. B must occupy one of the two end counters. C must not sit next to B. E must occupy the counter immediately to the right of C. F must not occupy an end counter, and F must not be adjacent to A. No two officers share a counter. The name plates are to be painted tonight, so the head wants one arrangement that satisfies every condition at once, together with a written justification that no other arrangement is possible. A trainee points out that if any one of the conditions is misrecorded, the whole row changes, and the head wants to know exactly how much the answer depends on each condition before he signs the order.

1. Fix the counter of every officer and show why no other arrangement satisfies all five conditions.

Read the worked answer

The key pair is 'A exactly three to the left of D'. The only possible slot pairs are (A, D) = (1,4), (2,5) and (3,6). Test each.

(3,6): counter 6 is taken, so B must be at 1. That leaves 2, 4 and 5 for C, E and F, but no two of these are adjacent, so 'E immediately right of C' cannot be met. Reject.

(2,5): if B is at 6, C and E must take 3 and 4, which forces F to 1 - adjacent to A at 2. Reject. If B is at 1, C and E take 3 and 4 and F lands on 6, an end counter. Reject.

(1,4): B must be at 6, the only remaining end. C and E must be 2 and 3 in that order, leaving F at 5 - not an end counter, and three places away from A. Every condition holds.

The only valid row is 1-A, 2-C, 3-E, 4-D, 5-F, 6-B. The reasoning is elimination, not recall: each rejected branch fails on one named condition, which is what the branch head needs to see on paper.

2. The trainee notices that the condition 'B must occupy one of the two end counters' was copied from an older circular and may not apply. If that condition alone is dropped, is the arrangement still unique?

Read the worked answer

No. Keep (A, D) = (1,4), (2,5) or (3,6) and re-test with B free, but F still barred from end counters and from A's neighbour.

(1,4): counters 2, 3, 5, 6 go to B, C, E, F. C-E adjacent is either 2-3 or 5-6. With C=2, E=3, F cannot be 6 (end), so F=5 and B=6 - the original row. With C=5, E=6, B and F take 2 and 3; F cannot be 2 (next to A), so F=3, B=2. This gives A-1, B-2, F-3, D-4, C-5, E-6, and C at 5 is not next to B at 2.

(2,5): leftover 1, 3, 4, 6. C-E can only be 3-4, leaving 1 and 6 for B and F, both end counters. F fails. Reject.

(3,6): leftover 1, 2, 4, 5. C-E is 1-2 or 4-5. Taking C=1, E=2, B and F take 4 and 5; F cannot be 4 (next to A at 3), so F=5, B=4. This gives C-1, E-2, A-3, B-4, F-5, D-6. The 4-5 option leaves F on 1 or 2, both illegal.

Three rows now survive, so the rule must be confirmed before the name plates are painted.

3. Suppose the branch instead relaxes only the condition 'E immediately to the right of C' to 'E is somewhere to the right of C', keeping all other conditions including B at an end. How many arrangements survive, and which condition now does the real work of narrowing the field?

Read the worked answer

Four arrangements survive.

With B at an end and F not at an end and not beside A:

  • (A,D) = (1,4), B = 6, remaining 2, 3, 5. If F = 3, C and E are 2 and 5 with E to the right, giving A-1, C-2, F-3, D-4, E-5, B-6. If F = 5, C and E are 2 and 3, giving A-1, C-2, E-3, D-4, F-5, B-6 (the original row).
  • (A,D) = (2,5), B = 1: F cannot be 1 (end) or 3 (beside A), so F = 4, C = 3, E = 6 โ†’ B-1, A-2, C-3, F-4, D-5, E-6.
  • (A,D) = (2,5), B = 6: F = 4 again, C = 1, E = 3 โ†’ C-1, A-2, E-3, F-4, D-5, B-6.
  • (A,D) = (3,6) forces B = 1 and C = 2, which puts C beside B, so this branch dies.

The condition doing the real work is now 'C is not next to B': every candidate in the (3,6) branch collapses on it. The lesson is that dropping an adjacency requirement costs more solutions than dropping an ordering requirement.