There are `2n` guests at a dinner party. Supposing that eh master and mistress of the house have fixed seats opposite one another and that there are two specified guests who must not be placed next to one another, show that the number of ways in which the company can be placed is `(2n-2!)xx(4n^2-6n+4)dot`

Explanation

Excluding the two specified guests, (2n-2) persons can be divided into two gropus, one containing n and the other containing n-2 in (2n-2)!/[n!(n-2)!] ways and can sit on either side of master and mistress in 2! Ways and can arrange themselves in n!(n-2)! ways. Now, the two specified guests where n-2 guests are seated will have (n-1) gaps and can arrange themselves in 2! ways. Hence, the number of ways when `G_(1)G_(2)` will always be together is

`((2n-2)!)/(n!(n-2)!)2!n!(n-2)!(n-1)xx2!`
`=4(n-1)xx(2n-1)!`
Hence, the number of ways when `G_(1)G_(2)` are never together is
`(2n)!=4(n-1)(2n-2)!`
`=(2n-2)![2n(2n-1)-4(n-1)]`
`=(2n-2)![4n^(2)-6n+4]`