Question

Six persons A, B , C, D, E and F are to be seated at a circular table . Find 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.

• A. $18$
• B. $27$
• C. $81$
• D. $9$

Answer 1

The correct option is A $18$

Say $B$ on right side of $A$

$DorC$ will be right of $B$

This can be done in $2!$ ways. Remaining three places can be filled in $3!$ ways

$=2!\times 3!=12$ ways.

Say $C$ on right of $A$

$B$ can be filled in $3$ places. (B next should be D).

$B$ immediately next is $D$

${}^{\prime }{A}^{\prime }{,}^{\prime }{C}^{\prime }{,}^{\prime }{B}^{\prime }{,}^{\prime }{D}^{\prime }$ are filled in 3 ways. Remaining two places in 2 ways.

Total $=6$ ways

Total number $=12+6=18$ ways.