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 the order
: AB or AC
When B has C or D to his right we have the order
: BC or BD
Taking these two possibilities together, we must
have ABC or ABD or AC and BD
For ABC, D, E, F to arrange around a circle no.
of ways $=3!=6$
For ABD, C, E, F to arrange around a circle no.
of ways $=3!=6$
For AC, BD, E, F to arrange around a circle no.
of ways $=3!=6$
$\therefore$ Total no. of ways $=6+6+6=18$