In how many ways can we seat 6 people around a round table if Fred and Gwen insist on sitting opposite each other?  (Two seatings are considered equivalent if one is a rotation of the other.)
There are 6 choices of seats for Fred to sit in.  Once Fred is seated, then Gwen must sit opposite him.  This leaves 4 people to place in the four remaining seats, which can be done in $4!$ ways.  However, we must divide by 6 to account for the 6 rotations of the table.  So the number of arrangements is $\dfrac{6 \times 1 \times 4!}{6} = 4! = \boxed{24}$.  Alternatively, we could start by fixing the table around Fred, thus removing the rotation.  There is 1 option for Gwen's seat, since she must sit across from him.  This leaves 4 people to place in four unique seats, so the number of arrangements is $4! = \boxed{24}$.