Question

Six persons A, B, C, D, E, F, are to be seated at a circular table. In how many ways antis be one if A should have either B or C on his and B must always have either C or D on his right.

Answer 1

Let the seat occupied by A be numbered as 1 and the remaining 5 seats be numbered as 2, 3,4,5,6 in anticlockwise direction. There arise two cases :
Case I : B is on right of A, i.e., at number 2.
Then, seat number 3 can be occupied by C or D in `.^(2)C_(1)` ways and remaining 3 persons can have remaining 3 seats in 3! ways. Hence, the number of arrangements in this case is `2xx6=12`.
Case II : C is on the right of A. i.e., at number 2.
Then, Bcan occupy any seat from number and 2 seats, which can be occupied in 2! ways. Hence, the number of arrangements in this case is `.^(3)C_(1)xx2!=6`. These cases are exclusive. So by the sum rule, the total number of arrangements is 12+6=18.