Seating four of ten guests at a round table

From ten people, four are chosen to sit at a round table. Two arrangements count as the same if every seated person has the same neighbour on their left and the same neighbour on their right. How many different arrangements are there?

$1260=\binom{10}{4}\cdot3!$. Choose the four diners in $\binom{10}4=210$ ways. Seatings that differ only by a rotation keep everyone's left and right neighbours, so fix one person and order the other three: $3!=6$. (Reflections swap left and right, so they are different.)