How many ways to seat 9 couple around a round table 
*

*You are a host/hostess at your local Applebee’s. You are seating a
group consisting of 9 couples at a round table. 
A)In how many different ways can you do this, provided that each couple will sit
together, and all that you care about is their position relative to
one another?  
B)What is the probability that Al doesn’t end up
within two seats of Ricky, AND Beth doesn’t end up within two seats
of Charlene?
Used a wrong tag before of order-statistics. Sorry about that. 
 A: For the number of arrangements, imagine that one of the chairs is a throne, and the Queen is one of the group at Applebee's. She sits down first, of course, on the throne.  Her Consort has $2$ choices of chair. Now let us seat the other couples one at a time, counterclockwise from the Queen-Consort pair. 
The person chosen to occupy the chair immediately counterclockwise from the royal pair can be chosen in $16$ ways. Now the occupant of the next chair is determined. The person chosen to occupy the next chair after that can be chosen in $14$ ways, and then the occupant of the chair after that is determined. And so on. 
Multiply. We get $(2)(16)(14)(12)\cdots (4)(2)$. This can also be written as $(2^9)(8!)$. 
A: Another approach to this counting problem is to think of each couple as "single" object, counting the number of ways the single objects can be arranged, and then multiplying by the number of equivalent ways the couples can be arranged as "compound" objects.
In particular, there are $9$ couples (considered as "single objects"), and they can be arranged around the table in $8!$ ways.  (Make sure you understand why this is so)
Now we consider the couples as "compound" objects, consisting of persons $A$ and $B$.  Each couple can be arranged in 2 ways:  $AB$ or $BA$.  Since there are $9$ couples, there are $2^9$ choices.
So we get $2^9 \cdot 8!$ arrangements.
And yes, this generalizes to arbitrary $n$:  if there are $n$ couples, they can be arranged in $2^n (n-1)!$ ways.
