Question

In how many ways can 12 distinct people be arranged around a circular table, if exactly two of them must always have at least one person between them?

• A. \(9 \times 9!\)
• B. \(9 \times 10!\)
• C. \(10 \times 10!\)
• D. \(8 \times 9!\)

Hint:

Subtract the arrangements where the two people sit adjacent from the total circular arrangements.

Answer 1

The Correct Option is B

Step 1: The total number of ways to seat 12 distinct people around a circular table, with rotations considered identical, is \((12-1)! = 11!\).

Step 2: Let the two special people be \(A\) and \(B\). Count the arrangements in which \(A\) and \(B\) sit next to each other, since these must be excluded. Treat \(A\) and \(B\) as a single block. This block together with the remaining 10 people gives 11 units to arrange in a circle, in \((11-1)! = 10!\) ways. Within the block, \(A\) and \(B\) can be ordered in \(2\) ways, so the number of arrangements with \(A, B\) adjacent is \(2 \times 10!\).

Step 3: The number of arrangements where \(A\) and \(B\) have at least one person between them is the total minus the adjacent count: \(11! - 2 \times 10! = 11 \times 10! - 2 \times 10! = 9 \times 10!\).

\[\boxed{9 \times 10!}\]