Question

There are 2n guests at a dinner party. Supposing that the master and the mistress have fixed seats opposite to one another and there are two specified guests who must not be placed next to one another, show that the number of was in which the company can be placed is $\left( {2{\text{n}} - 2} \right)! \times \left( {4{{\text{n}}^2} - 6{\text{n}} + 4} \right)$.

Answer 1

Hint: In this question the concept of permutations and combinations will be used. We need to apply logic and general algebra to proceed through the problem. We will first find all the cases in which the two guests cannot be placed next to one another, and then add them. In order to select r unique things from a total number of n things, the number of ways are-
${}_{}^{\text{n}}{\text{C}}_{\text{r}}^{} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!{\text{r}}!}}$

Complete step-by-step answer:
Let M and N 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 X and Y.

Now, we will put X in a position adjacent to M$(a_1)$. Now, Y can be placed everywhere except $a_2$, and the place occupied by X. Let us place Y at the place $a_3$. The rest of the $2n - 2$ places can be arranged in $(2n - 2)!$ ways. Similarly, Y can be placed in $2n - 2$ such places. Hence, the total number of ways in which guests are placed when X is at $a_1$ are-
(2n - 2)(2n - 2)!
Similarly, X can be placed in four such places adjacent to M and N, so the total number of ways for this are-
4(2n - 2)(2n - 2)!

When X is placed in any of the remaining $2n - 4$ places, Y cannot be placed in the two places adjacent to X. So, the number of places in which Y can be placed are $(2n - 3)$. The remaining of the $(2n - 2)$ guests can be placed in $(2n - 2)!$ ways. So the total number of ways in which all the guests can be arranged are-
$(2n - 4)(2n - 3)(2n - 2)!$

We will add these two separate cases to get the total number of ways in which the guests can be arranged. This is given by-
$4(2n - 2)(2n - 2)! + (2n - 4)(2n - 3)(2n - 2)!$
$ = \left( {2{\text{n}} - 2} \right)! \times \left( {8{\text{n}} - 8 + 4{{\text{n}}^2} - 14{\text{n}} + 12} \right)$
$ = \left( {2{\text{n}} - 2} \right)! \times \left( {4{{\text{n}}^2} - 6{\text{n}} + 4} \right)$
This is the number of ways in which the company can be placed.


Note: There is no direct method to solve this problem. We need to form the equations according to the question, and simplify it using general algebra. One common mistake is that students often forget the formula for combinations and permutations, which should be remembered. Also, we should keep in mind the arrangement of the guests is circular, and not linear.