Q.In how many ways can 6 distinct people be seated around a circular table, if

Circular seating removes the usual row-arrangement's fixed starting point, so the count uses $(n-1)!$, halved further if clockwise and anticlockwise are treated as the same arrangement. …

(i) For $n$ distinct people seated around a circular table where clockwise and anticlockwise arrangements are counted as different, the number of distinct circular arrangements is $(n-1)!$. With $n=6$: $(6-1)! = 5! = 120$.

(ii) If a clockwise arrangement and its mirror-image anticlockwise arrangement are considered the same seating (as when the table can effectively be viewed/flipped from either side), the count from (i) is halved: $\dfrac{5!}{2} = \dfrac{120}{2}=60$. …

Using the row-permutation formula $6!=720$ directly, forgetting that a circular arrangement has no fixed starting point and so must use $(n-1)!$ instead of $n!$; also forgetting to halve the count i …