Question

A party of n persons sit at a round table. What are the odds against two specified persons sitting next to each other?

• A. $(n-3):(n-1)$
• B. $(n-1):(n-3)$
• C. $(n-2):(n-3)$
• D. $(n-1):(n-2)$

Answer 1

The correct option is A $(n-3):(n-1)$

Total no. of ways in which n people can sit in a round table$=(n-1)!$

If we take two specified persons as one group there will be $(n-1)$ entities.

Total no. of ways there $(n-1)$ entities can be arranged in a round table$=(n-2)!$

Also, two person can sit in $2!$ ways

Thus, P(two specified people can sit together in n-people round table)$=\frac{2(n-2)!}{(n-1)!}=\frac{2!}{(n-1)}$

Since the question acts the probability against it is $=1-\frac{2}{(n-1)}=\frac{n-3}{n-1}$.