There are $n$ couples, and they will sit at random at a round table. What is the probability that there will be exactly one couple sitting next to each other?
Here is my attempt: I know how to solve the following problem by using the inclusion-exclusion principle.
There are $n$ couples, and they will sit at random at a round table. What is the probability that there will be no couple sitting next to each other?
Let $E_i$ denothe event that $i^{th}$ couple sits together. Then $$P(\hbox{no couple sits together})=1-P(\cup_{i=1}^n E_i)=1-\left[nP(E_1)-\binom{n}{2}P(E_1\cap E_2)+\binom{n}{3}P(E_1\cap E_2\cap E_3)\cdots (-1)^{n-1}P(E_1\cap E_2\cap \cdots \cap E_n)\right]$$
I know how to calculate the probabilities of the above events. For instance, $$P(E_1)=\frac{2(2n-2+1-1)!}{(2n-1)!}, \quad P(E_1\cap E_2)=\frac{2^2(2n-4+2-1)!}{(2n-1)!} $$
Now, we can go from this solution to the original question. The original question can be considered as follows: there are $n-1$ couple and one single person (corresponding to the couple sitting together). They will sit at random at a round table. What is the probability that there will be no couple sitting next to each other?
For this case, $$P(E_1)=\frac{2.2(2n-4+2-1)!}{(2n-2+1-1)!}$$.
Am I correct?
As a follow-up question I would like to ask the following:
There are $n$ couples and they will sit at random at a round table. Define the random variable $X_n=$the number of couples sitting next to each other. What can be say about the asymptotic growth of $E(X_n)$ and $Var(X_n)$ as $n\to \infty$ and also the limiting distribution of $\frac{X_n-E(X_n)}{\sqrt{Var(X_n)}}$?