TITLE: n married couples, are randomly seated at a round table. Find probability for k couples to sit together.
QUESTION [7 upvotes]: A total of 2n people, consisting of n married couples, are randomly seated at a round table. Find the probability for k of the couples (k<n) to be seated together.
Let us first find the probability for 1 of the couples to be seated together.
Once the 2 persons are seated somewhere, the remaining $2n-2$ can be seated in $(2n-2)!$ ways, plus that the 2 members of the couple can be also seated in $2!$ ways. On the other hand, 2n people in a circular arrangement can be seated in $(2n-1)!$ ways, therefore the probability is
$\frac {2!(2n-2)!}{(2n-1)!} = \frac{2}{(2n-1)}$.
Now let's consider a second couple sitting together: Once the 1st couple has seated, we have $2n-2$ free seats, the 1st couple (which we consider as one unit) and a 2nd couple (also as one unit), so $2n-4+1$ can be arranged in $(2n-3)!$ ways, and the 2 couples can also be arranged in $2!.2!$ ways. So the probability for 2 couples to sit together (somewhere in the table) is $\frac{2!.2!.(2n-3)!}{(2n-1)!}$.
For the $k_{th}$ couple, the probability is $\frac{2!.2!.2!...2!.k!.(2n-2k-1)!}{(2n-1)!}$.
Is it correct?
I know this is a very common problem but haven't found this variation (with k couples) anywhere.
Can you help me? Thank you!

REPLY [2 votes]: Firstly, consider the $k$ couples as occupying one "double" seat each in $2$ ways, so there are $2n-2k +k = 2n-k\;$ seats available, and number of ways of seating them with $k$ couples together $= 2^k\cdot(2n-k-1)! $
And w/o restrictions, there are $= (2n-1)!\;$ways
hence $Pr = \frac{2^k\,\cdot\,(2n-k-1)!}{(2n-1)!}$