Gate probability

Question

A party of n persons take their seats at random at a round table,then the probability that two specified person do not sit together is

• A. $\left(\dfrac{2}{n-1}\right)$
• B. $\left(\dfrac{n-3}{n-1}\right)$
• C. $\left(\dfrac{n-2}{n-1}\right)$
• D. $\left(\dfrac{1}{n-1}\right)$

Answer 1

The number of ways to seat $n$ people at a round table is $(n-1)!$

Two people can sit together in $2!$ ways

$P1P2 $ or $P2P1$

seat them anywhere, and seat the rest in $(n-2)!$ ways

So, the total number of ways for two people together and rest is $2*(n-2)!$

P(2 sit together ) = $2*(n-2)! / (n-1)!$

P(not 2 sit together) = $1 -  2*(n-2)! / (n-1)! $

                                 = $1 -  2*/ (n-1) $

                                 = $ n-3/n-1$

Comments

I have a query though your answer is correct.
"Why didn't you choose two people initially?" i.e. like
Step-1: Choose Two persons in C(n,2) ways.
Step-2: Further whatever is done in above solution.

Answer 2

(n-2) persons excluding the two person can be arranged in (n-3)! ways 2 persons can be seated among (n-2) persons in $\binom{n-2}{2}*2!$ therefore prob of  two person can not seat together ((n-3)!* $\binom{n-2}{2}*2!$)/(n-1)!

on solving we get (n-3)/(n-1)