Question

Six persons $A,B,C,D,E,F$ are to be seated at a circular table. If $A$ should have either $B$ or $C$ on his immediate right and $B$ must always have either $C$ or $D$ on his immediate right. Then the total number of possible arrangements is

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 immediate 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:
$={}^2{C}_1\times 3!=2\times 6=12.$

Case II: $C$ is on the immediate right of $A$ i.e., at number $2$.
Then, $B$ can occupy any seat from number $3$ or $4$ or $5$(not $6$). Then, $D$ must be on the right of $B$, so we are left with two persons and $2$ seats, which can be occupied in $2!$ ways.

Hence, the number of arrangement in this case is:
$={}^3{C}_1\times 2!=3\times 2=6.$

These cases are exclusive. So by sum rule, total number of arrangements is
$=12+6=18.$