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There are $10$ people sitting in a round table. Bob and Ben have to sit opposite from each other, and Jeff and John also have to sit opposite from each other. How many possible seatings are there?

I think it should be $9 \cdot 6!$, since you can pick Bob first, and then pick Ben, leaving $9$ spots for Jeff and John to be. Then you can arrange the rest of the people in $6!$ ways.

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    $\begingroup$ If you pick Bob's seat, Ben's seat is also forced to be across from him. Does that still leave 9 seats left? $\endgroup$ – WaveX Sep 7 '20 at 2:23
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Seat Bob. We will use him as our reference point. Ben must sit opposite to him. There are eight seats remaining, so there are eight ways to seat Jeff. John must sit opposite to him. The remaining six people may be seated in $6!$ ways as we proceed clockwise around the table from Bob. Hence, there are $8 \cdot 6!$ admissible seating arrangements.

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