Q.There are 10 persons among whom two are brothers. The total number of ways in which these persons can be seated around a round table so that exactly one person sits between the brothers, is equal to:
A) 2!×7! B) 2!×8! C) 3!×7! D) 3!×8!

A circular permutation is an arrangement of objects around a closed loop (a round table, a garland, a necklace) rather than in a straight row, and it is counted differently from a row-arrangement because a circular arrangement has no fixed starting point: rotating every object one seat around produces the SAME relative arrangement, since everyone still has exactly the same left- and right-neighbour as before. If $n$ distinct objects were instead arranged in a row, there would be $n!$ linear arrangements; but each single circular arrangement, when 'cut open' at any one of its $n$ positions, corresponds to exactly $n$ different linear arrangements (one for each place the cut could be made) that are all really the same circular seating. So the number of distinct circular arrangements of $n$ distinct objects is $\dfrac{n!}{n}=(n-1)!$ — equivalently, fix any one object's seat to remove the rotational duplicate-counting, and arrange the remaining $n-1$ objects relative to it in $(n-1)!$ ways. Two important refinements: if $m$ of the $n$ objects are identical, the count becomes $\dfrac{(n-1)!}{m!}$ (dividing out their internal rearrangements just as with row permutations). And if a clockwise seating and its exact mirror-image (anticlockwise) seating are considered to be the SAME arrangeme …