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

Concept: Number of circular arrangements of \(n\) distinct persons: \[\begin{aligned} (n-1)! \end{aligned}\] Required arrangements \[\begin{aligned} = \text{Total arrangements} - \text{Arrangements where the two particular delegates sit together} \end{aligned}\]

Step 1: Calculate total circular arrangements. \[\begin{aligned} (20-1)! = 19! \end{aligned}\]

Step 2: Calculate arrangements where the two delegates sit together. Treat the two delegates as one block. Then total units \[\begin{aligned} =19 \end{aligned}\] Circular arrangements: \[\begin{aligned} (19-1)! = 18! \end{aligned}\] The two delegates can interchange places in \[\begin{aligned} 2! =2 \end{aligned}\] ways. Hence, \[\begin{aligned} \text{Together} = 2\times18! \end{aligned}\]

Step 3: Find the required number of arrangements. \[\begin{aligned} 19! - 2\times18! \end{aligned}\] \[\begin{aligned} \boxed{ 19!-2\times18! } \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.