Enemy knights at a round table

Question

King Arthur summoned $2n$ knights to his court. He paired each knight to his sworn enemy so there are $n \geq 1$ pairs total. How many ways can the knights be seated at a round table so that no two enemies sit next to each other?

Answer 1

Let the people be {$A_{1,1},A_{1,2},A_{2,1},A_{2,2},A_{3,1},A_{3,2}, ... ,A_{n,1},A_{n,2}$}.
Where $A_{i,1}$ and $A_{i,2}$ are enemies.

$\therefore$ By the principle of inclusion-exclusion we can say that -
The required answer =$\sum_{i=0}^n(-1)^n\binom{n}{i}A_i$
Where $A_i$ stands for number of arrangements possible by having $i$ specific pairs of enemies together and we can choose those $i$ pairs in $\binom{n}{i}$ ways.

To find $A_i$ -

We can count each pair as only one person.Then we will have $(2n-i)$ people giving us $(2n-i-1)!$ arrangements. Then we can also switch places with each pair having $2^i$ ways.

$\therefore A_i = 2^i(2n-i-1)!$

$\therefore$ Required answer =$$\sum_{i=0}^n(-1)^n\binom{n}{i}(2^i)(2n-i-1)!$$

Answer 2

Sit the first person down anywhere which doesn't matter. Now to the right of him there are $2n-2$ people who can sit there since anyone but his enemy can sit there. Now to the right of that person there are $2n-3$ people who can sit there. Continue this and you should get the answer of $(2n-2)!$. However, you also need to consider the other direction where the arrangement goes counter clockwise. Therefore, it should be $2(2n-2)!$.