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 $18$

For these type of question we need to consider different cases.

One case with $A$, having $B$ to his right and $C$ to $B$ right. Now number of ways the rest can be arranged is $3!$

Now we take $A$ with $B$ in right but this time $D$ to $B$ right again the rest can be arranged in $3!$ ways.

For the last case we have $A$ and $C$ to his right. we are left with $E,F$and $BD$. (BD one entity as $B$ must be seated with $D$ as immediate right)

$\therefore$ again $3!$ ways

So total no of ways $=3\times 3!$

$=18$ ways.

Hence, the answer is $18.$