Question

There are ${\text{2n}}$ 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 $\left( {{\text{m}}\left( {{{\text{n}}^2}} \right) - {\text{k}}\left( {\text{n}} \right) + 4} \right)\left( {2{\text{n - 2}}} \right)!$. Find k and m.

Answer 1

Hint:
Here, let us assume that A and B are the two guests who cannot be placed next to one another so if their place is fixed then the seats left are $2{\text{n}} - 2$. Now, we will find all the cases in which the two guests cannot be placed next to one another and add them. For this we can use the concept of combination- to select r number of things out of n total number of things, we use the following formula-${}^{\text{n}}{{\text{C}}_{\text{r}}}$
And the formula of combination is given as-${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\frac{{n!}}{{r!n - r!}}$ Where n=total number of things and r = number of things to be selected.

Complete step by step solution:
Let M be the master and N be the mistress of the house. Now it is given that there are ${\text{2n}}$guests at a dinner party so there must be $2{\text{n}}$seats for the guests. Let the assigned seats be ${{\text{a}}_{\text{1}}}$,${{\text{a}}_2}$,…,${{\text{a}}_{{\text{2n}}}}$that represent $2{\text{n}}$ seats.
Now, there are two guests who must not be placed next to one other so let the two guests be A and B.
Now, if we put A in a position adjacent to M at ${{\text{a}}_{\text{1}}}$then B can be placed anywhere except ${{\text{a}}_{\text{1}}}$and${{\text{a}}_2}$as ${{\text{a}}_{\text{1}}}$ is already occupied by A and ${{\text{a}}_2}$is right next to the seat assigned to A.
So let the seat assigned to B be ${{\text{a}}_{\text{3}}}$.Then, the seats left are$2{\text{n}} - 2$.
 So these can be arranged in total number of ways=$\left( {2{\text{n}} - 2} \right)!$
$ \Rightarrow $ $2{\text{n}} - 2$guest can be placed in $2{\text{n}} - 2$seats when A is at ${{\text{a}}_{\text{1}}}$ in total number of ways=$\left( {2{\text{n}} - 2} \right)\left( {2{\text{n}} - 2} \right)!$
Similarly, A can be placed in four such places adjacent to M and N so all guests can be arranged in the total number of ways=$4\left( {2{\text{n}} - 2} \right)\left( {2{\text{n}} - 2} \right)!$-- (i)
When A is placed in any remaining $2{\text{n}} - 4$places then B cannot be placed in the two seats adjacent to A so the total number of seats in which B can be placed is $2{\text{n}} - 3$.
Now, the remaining $2{\text{n}} - 2$guest can be placed in total number of ways=$\left( {2{\text{n}} - 2} \right)!$
Now, all the guest can be arranged in total number of ways=$\left( {2{\text{n}} - 4} \right)\left( {2{\text{n}} - 3} \right)\left( {2{\text{n}} - 2} \right)!$-- (ii)
On adding eq. (i) and (ii), we get-
$ \Rightarrow $ The total number of ways the guests can be arranged=$4\left( {2{\text{n}} - 2} \right)\left( {2{\text{n}} - 2} \right)! + \left( {2{\text{n}} - 4} \right)\left( {2{\text{n}} - 3} \right)\left( {2{\text{n}} - 2} \right)!$
On taking the common terms out from the first and second term, we get-
$ \Rightarrow $ The total number of ways the guests can be arranged=$\left( {2{\text{n}} - 2} \right)!\left[ {4\left( {2{\text{n}} - 2} \right) + \left( {2{\text{n}} - 4} \right)\left( {2{\text{n}} - 3} \right)} \right]$
On solving the terms inside the bracket, we get-
$ \Rightarrow $ The total number of ways the guests can be arranged=$\left( {2{\text{n}} - 2} \right)!\left[ {{\text{8n}} - 8 + 4{{\text{n}}^2} - 6{\text{n}} - 8{\text{n + 12}}} \right]$
On simplifying, we get-
$ \Rightarrow $ The total number of ways the guests can be arranged=$\left( {2{\text{n}} - 2} \right)!\left[ {4{{\text{n}}^2} - 6{\text{n + 12}} - 8} \right]$
On further simplifying, we get-
$ \Rightarrow $ The total number of ways the guests can be arranged=$\left( {2{\text{n}} - 2} \right)!\left[ {4{{\text{n}}^2} - 6{\text{n + 4}}} \right]$
On comparing this result with $\left( {{\text{m}}\left( {{{\text{n}}^2}} \right) - {\text{k}}\left( {\text{n}} \right) + 4} \right)\left( {2{\text{n - 2}}} \right)!$, we get-

m=$4$ and k=$6$

Note:
Here, we have to remember that the arrangement of the guests is circular, not linear. Also, here we cannot solve this question directly using the formula. We have to form equations for each case following the condition given in the question and add all of them to make it easy to find the arrangement.