A round-table conference is to be held among 20 delegates belonging from 20 different countries. In how many ways can they be seated if two particular delegates are (i) always to sit together, (ii) never to sit together.

Explanation

(i) Let the 20 delegates be `D_(1),D_(2),D_(3)………….D_(20)`. Let 2 particular delegates `D_(1), D_(2)` sit together .They can do so in 2! =2 ways . Now supposing `D_(1) & D_(2)` together as one member, there are `(1,D_(3),D_(4),D_(5),……..D_(20))`=19 members to be seated on a round table in (18)! ways. Hence total number of ways in which `D_(1)` and `D_(2)` sit together =`2(18)!`
(Iii) Let `D_(1) , D_(2)` never sit together , consider 19 delegates excluding `D_(1)`, they are sit on a round table in (18)! ways . For each sitting arrangement there are 19 places in between the delegates of which 2 positions adjacent to `D_(2)` cannot be occupied by `D_(1)` so for each arrangement of `D_(2),D_(3),D_(4).........D_(19),D_(20)` on the table there are 17 possitions which `D_(1)` can occupy . Hence by multiplication principle total number of ways in which `D_(1) , D_(2)` can never sit together =17(18)!

Second method
Total number of arrangement of 20 delegates on a round table without any restriction =(20-1)!=(19)! ways.
Now two particular delegates can sit together =2!(18)!
`therefore ` Number of ways in which two particular delegates never sit together =(19)!-2(18)!=(19-2)(18)!=17(18)!