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, Let the number of ways in which the company can be placed be $\left(m\right(n^2)-k(n)+4)(2n-2)!$.find k-m ?

Answer 1

Let $M$ and $M^{\prime }$ represent seats of the master and mistress respectively, and 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 $P$ are $a_1$, and $Q$ at any position, other than $a_2$, say at $a_3$; then remaining $2n-2$ guests can be arranged in the remaining $(2n-2)$ positions in $(2n-2)!$ ways.
Hence there will be altogether $(2n-2)(2n-2)!$ arrangements of the guests when $P$ is at $a_1$.
The same number of arrangement when $P$ is at $a_n$ or $a_{n+1}$ or $a_{2n}$. Hence for these positions $(a_1,a_n,a_{n+1},a_{2n})$ of $P$ there are altogether
$4(2n-2)(2n-2)!$ ways $....\left(1\right)$
If $P$ is at $a_2$ 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 excepting 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 $....\left(2\right)$
Hence from $\left(1\right)$ and $\left(2\right)$, the total no. of ways of arranging the guests
$=4(2n-2)(2n-2)!+(2n-4)(2n-3)(2n-2)!$
$=(4n^2-6n+4)(2n-2)!$