Question

Six persons $A,B,C,D,E$ and $F$ are to be 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

• A. $36$
• B. $12$
• C. $24$
• D. $18$

Answer 1

The correct option is D $18$

When $A$ has $B$ or $C$ to his right, we have $AB$ or $AC$
When $B$ has $C$ or $D$ to his right, we have $BC$ or $BD$.
Thus, we must have $ABC$ or $ABD$ or $AC$ and $BD$.
For $ABC$, $D,E,F$ on a circle number of ways$=3!=6$
For $ABD$, $C,E,F$ on a circle number of ways$=3!=6$
For $AC,BD,E,F$ the number of ways$=3!=6$
Hence, the required number of ways $=18$