Question

A party of n persons sit at a round table, find the odds against two specified individuals sitting next to each other.

Answer 1

No. of ways in which $n$ people can seat in round table$=(n-1)!$
Consider two specified as one group then there will be $(n-1)$ persons.
So, now no. of ways $(n-1)$ can seat in round table$=(n-2)!$
No. of two specified can seat$=2!$ ways
Total no. of ways $\Rightarrow (n-2)!2!$
Required probability$=\frac{(n-2)!2!}{(n-1)!}=\frac{2}{n-1}$
So probability against = $\frac{1-requiredprobability}{requiredprobability}$
$\Rightarrow \frac{1-\frac{2}{n-1}}{\frac{2}{n-1}}\Rightarrow \frac{\frac{n-3}{n-1}}{\frac{2}{n-1}}$

$=\frac{n-3}{2}$