Question

20 delegates from 20 countries sit in a circle such that two particular delegates never sit together. In how many ways can they be seated?

• A. \(20!-2\)
• B. \(19!-2\times18!\)
• C. \(19!-18!\)
• D. \(17!\times18\)

Hint:

For circular permutations: \[ (n-1)! \] When two persons must stay together, treat them as one block and multiply by \(2!\).

Answer 1

The Correct Option is B

Step 1: Recall the circle rule.
When $n$ different people sit around a circle, the number of seatings is $(n-1)!$ because rotations count as the same arrangement.

Step 2: Count all seatings of the 20 delegates.
With $n = 20$, the total is $(20-1)! = 19!$.

Step 3: Use the trick of removing the bad cases.
We want the two special delegates apart. Easier to count the seatings where they ARE together, then subtract from the total.

Step 4: Count the seatings where the two sit together.
Tie the two special delegates into one block. Now we are arranging $19$ items (the block plus $18$ others) in a circle, giving $(19-1)! = 18!$ ways.

Step 5: Account for the two inside the block.
Inside the block, the two can swap places in $2$ ways. So together-seatings $= 18! \times 2 = 2 \times 18!$.

Step 6: Subtract to get them apart.
Required $= \text{total} - \text{together} = 19! - 2 \times 18!$. \[ \boxed{19! - 2 \times 18!} \]