Question

There are $2n$ guests at a dinner party. Supposing that the 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 $(2n-1)!\times (4n^2-6n+4)$.

Answer 1

Excluding the two specified guests, $2n$ persons can be divided into two groups one containing $n$ and the other containing $n-2$ in $\frac{(2n-2)!}{[n!(n-2)!]}$ ways and can sit on either of master and mistress is $2!$ ways and can arrange themselves in $n!(n-2)!$.

Now, the two specified guests where $n-2$ guests are seated will have $n-1$ gaps and can arrange themselves in $2!$ ways.

The number of ways when $G_1{G}_2$ will always be together is
$\frac{(2n-2)!}{n!(n-2)!}2!n!(n-2)!(n-1)\times 2!$

$=(2n-2)!4(n-1)$

Hence, the number of ways when $G_1{G}_2$ are never together is
$\left(2n\right)!-4(n-1)(2n-2)!=(2n-2)!\left[2n\right(2n-1)-4(n-1\left)\right]$

$=(2n-2)![4n^2-6n+4]$.