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

Let the M and `M^(')` represent seats of the master and mistress respectively annd let `a_(1),a_(2),a_(3), . .,a_(2n)` represent the 2n seats.

Let the guests who must not be placed next to one another be called P and Q.
Now, put P at `a_(1)` and Q at any position, other than `a_(2)`, say at remaining (2n-2) positions in (2n-2)! ways. hence, there will be altogether (2n-2)(2n-2)! arrangments of the guests, when P is at `a_(1)`.
the same number of arrangements when P is at `a_(n)` or `a_(n+1)` or `a_(2n)`. thus, for these positions `(a_(1),a_(n),a_(n+1),a_(2n))` or P, there are altogether `4(2n-2)(2n-2)!` ways.
If P is at `a_(2)`, then there are altogether `(2n-3)` positions for Q. hence, there will be altogether `(2n-3)(2n-2)!`
arrangements of the guests, when P is at `a_(2)`.
the same number of arrangements can be made when P is at any other position exception the four positions.
`a_(1),a_(n),a_(n+1),a_(2n)`.
Hence, for these (2n-4) positions of P, there will be altogether
`(2n-4)(2n-3)(2n-2)!` arrangements of the guests . . . (ii)
Hence, from Eqs. (i) and (ii), the total number of ways of arranging the guests
`=4(2n-2)(2n-2)!+(2n-4)(2n-3)(2n-2)!`
`=(4n^(2)-6n+4)(2n-2)!`.