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Question: There are \({\text{2n}}\) guests at a dinner party. Supposing that the master and mistress of the ho...

There are 2n{\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 (m(n2)−k(n)+4)(2n - 2)!\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.

Explanation

Solution

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 2n−22{\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-nCr{}^{\text{n}}{{\text{C}}_{\text{r}}}
And the formula of combination is given as-nCr{}^{\text{n}}{{\text{C}}_{\text{r}}}=n!r!n−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 2n{\text{2n}}guests at a dinner party so there must be 2n2{\text{n}}seats for the guests. Let the assigned seats be a1{{\text{a}}_{\text{1}}},a2{{\text{a}}_2},…,a2n{{\text{a}}_{{\text{2n}}}}that represent 2n2{\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 a1{{\text{a}}_{\text{1}}}then B can be placed anywhere except a1{{\text{a}}_{\text{1}}}anda2{{\text{a}}_2}as a1{{\text{a}}_{\text{1}}} is already occupied by A and a2{{\text{a}}_2}is right next to the seat assigned to A.
So let the seat assigned to B be a3{{\text{a}}_{\text{3}}}.Then, the seats left are2n−22{\text{n}} - 2.
So these can be arranged in total number of ways=(2n−2)!\left( {2{\text{n}} - 2} \right)!
⇒\Rightarrow 2n−22{\text{n}} - 2guest can be placed in 2n−22{\text{n}} - 2seats when A is at a1{{\text{a}}_{\text{1}}} in total number of ways=(2n−2)(2n−2)!\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(2n−2)(2n−2)!4\left( {2{\text{n}} - 2} \right)\left( {2{\text{n}} - 2} \right)!-- (i)
When A is placed in any remaining 2n−42{\text{n}} - 4places 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 2n−32{\text{n}} - 3.
Now, the remaining 2n−22{\text{n}} - 2guest can be placed in total number of ways=(2n−2)!\left( {2{\text{n}} - 2} \right)!
Now, all the guest can be arranged in total number of ways=(2n−4)(2n−3)(2n−2)!\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(2n−2)(2n−2)!+(2n−4)(2n−3)(2n−2)!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=(2n−2)![4(2n−2)+(2n−4)(2n−3)]\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=(2n−2)![8n−8+4n2−6n−8n + 12]\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=(2n−2)![4n2−6n + 12−8]\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=(2n−2)![4n2−6n + 4]\left( {2{\text{n}} - 2} \right)!\left[ {4{{\text{n}}^2} - 6{\text{n + 4}}} \right]
On comparing this result with (m(n2)−k(n)+4)(2n - 2)!\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=44 and k=66

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.