Question

Six persons $A,B,C,D,E,F$ are to be seated around a circular table facing the table. If $A$ should have either $B$ or $C$ on his/her adjacent side and $B$ must have either $C$ or $D$ on his/her right, then the number of 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 clockwise direction. There arise three cases :

Case I: $B$ is on right of $A$, i.e., at number $2$.
Then seat number $3$ can be occupied by either $C$ or $D$.
If $C$ occupies seat number $3$, then other $3$ can be arranged in $3!$ ways.
If $D$ occupies seat number $3$, then $C$ can sit on $4$ or $5$ and other $2$ can be arranged in $2!$ ways.
Number of ways $3!+2\times 2!=10$


Case II: $C$ is on the right of $A$ i.e., at number $2$.
Then, $B$ can occupy any seat from number $3$ or $4$ or $5$. 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.


Number of ways $3\times 2!=6$

Case III: $C$ is on the left of $A$ i.e., at number $2$.
(Note: $B$ can't be on the left of $A$)
Then, if $B$ occupies seat number $5$, then other $3$ can be arranged in $3!=6$ ways
If $B$ occupies seat number $4$ or $3$, then $D$ will be on $5$ or $4$ respectively. Other $2$ can be arranged in $2!$ ways.


Number of ways $6+2\times 2!=10$
Total number of ways from all the three cases is $10+6+10=26$