Ways to seat 6 couples in a circle with 14 chairs. Suppose there is a round table with $14$ chairs. How many ways can I seat $6$ couples around the table if I require partners of each couple to seat together?
My answer is $$\Big(\frac{13!}{8!}\Big)\cdot(2^6)$$.
I treat each couple as $1$ object. So there are $6$ objects and $8$ "ghosts". Then for each couple, they can be seated in $2$ ways. Is this correct?
 A: I would say:


*

*take any of the 14 chairs to be empty

*there are 7 positions allowed for the second empty chair (0,2,4,..,12 positions between them)

*for symmetry, divide by 2, so we have 49 configurations. Your answer does not have factor 49, so I think it is incorrect.

*permute the pairs (6!)

*order within the pairs (2^6)

*multiply.
A: Let the couples be $U$, $V$, $W$, $X$, $Y$, $Z$.  Let $E$ refer to an empty seat.  Let the chairs be numbered $1$–$14$.  Seat couple $U$ in chairs $1$ and $2$.  To fill chairs $3$–$14$, arrange the letters of the word $VWXYZEE$, which can be done in $\frac{7!}{2!}$ ways, and place the couples and empty seats accordingly.  Now decide which member of each couple sits to the right and which to the left.  This can be done in $2^6$ ways.
If the circularity of the table is meant to imply that a cyclic permutation of the occupants of the chairs gives an equivalent arrangement, than we're done.  (Requiring that couple $U$ occupy chairs $1$ and $2$ has given us a unique representative of each equivalence class.)  Otherwise there is an additional factor of $14$ to account for all possible positions of couple $U$.
Note that the first of these answers agrees with that of fawningflagellum, the second with that of Pieter21.
A: Cut down on the ghosts!
The answer is 7!2^5. Think about it!
