Question

Six persons $A,B,C,D,E$ and $F$ are to seated at a circular table. The number of ways this can be done, if $A$ must have either $B$ or $C$ on his right and $B$ must have either $C$ or $D$ on his right, is:

Answer 1

There are $6$ persons $A,B,C,D,E,F.$

$B$ or $C$ must be to the right of $A$ whereas $C$ or $D$ must be to the right of $B.$

Case $i)$ $B$ is to the right of $A.$

$A\xi B$ can be placed seat, since it is circular table it doesn't matter where we start.

$C$ or $D$ can be the right of $B={}^2{C}_1$

Remaining $3$ can be seated in $3!$ ways.

Total $={}^2{C}_1\times 3!$

Case $ii)$ $C$ is to right of $A.$

$A,C$ are fixed . we have $4$ places remaining. $B$ can be placed only in $3$ of these places,

if $B$ is placed to the seat which is left of $A$ then condition that $C,D$ should be to right of $B$ will not be satisfied.

$\therefore 3$ ways.

Now $D$ must be to right of $B$ in all $3$ ways, remaining $2$ can be seated in $2!$ ways.

Total $={}^3{C}_1\times 2!=6$

$\therefore$ Required $=12+6=18$ ways.

Hence, the answer is $18.$