In how many ways can 8 persons be seated at a round table so that all shall not have the same neighbour in any two arrangement?
By using the formula $(n-1) !$ (Mention in Solution-1)
So 8 persons can be arranged by $7 !$
Now each person have the same neighbour in the clockwise and anticlockwise arrangement
Total number of arrangement are $(7 !) / 2=2520$