Probablity that 3 husbands sit next to their wives round a circular table There are 3 couples sitting randomly round a 6-seater circular table. What is the probability that all the husbands and wives sit next to each other?
My attempt:
First wife, say, takes any of the six seats.  That leaves 2/5 seats where her husband can sit next to her.
Second wife, say, can take any of the four remaining seats.  There is then only 1 seat out of the remaining 3 where her husband can sit next to her AND leave two empty adjacent seats for the last couple.
So the answer is 2/5 * 1/3 = 2/15.
 A: The general case:
Consider $n$ groups of $k$.
Because the table is round there is a cyclic symmetry.
The total number of permutations is therefore $(nk-1)!$.
Each group has $k!$ permutations.
The total number of permutations for the groups is $(n-1)!$.
So we get
$$
\frac{k!^n (n-1)!}{(nk-1)!}.
$$

Case $n=3$ and $k=2$ gives
$$
\frac{2!^3 (3-1)!}{(2\cdot3-1)!} = \frac{2}{15}.
$$
A: Answer:
6 people can sit around a circular table in $(6-1)! = 5!$ ways
Fix each couple as one. They can be permuted in 2 ways.  There are three couples, so each one of them can be seated in $2^3$ ways as couple.  Now the three couples are three people and they can be placed in around the circular table in $2!$ ways.
Now the required probability  $= \frac{2^3.2}{5!} = \frac{2}{15}$ 
This is one way you can solve it.
Thanks
Satish
A: I think you might do too the following: 
6 persons can sitting randomly 6! ways.
3 couples (whole) can sitting randomly 3!*2 ways. Each of the pairs may sit in two ways (wife, husband or husband, wife) -  that's $2^3=8$ ways.
So I would say that the  answer is $\frac{3!\cdot 2\cdot 8}{6!}=\frac{2}{15}$
$\color{red}{So\, the\, solution\, is\, not\, correct - see\, comments.}$
A: So 2/15 if they all sit randomly, let’s say like with blind folds on. The actual probability depends on their wants and priorities. Does A actually not want to sit next to his wife A1? Does B1 want sit next to C? And so on. The probability depends entirely on the individuals involved. And yes, this particular answer was partly dependant on that I just could not do the math. :-)
