Permutation and CombinationHard
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:
Options
A.36
B.12
C.24
D.18
Solution
Case-I If B is right on A
Subcase -I C is right on B
then no. of ways = (4 - 1)! = 6
Subcase- II If D is right on B
then no. of ways = (4 - 1)! = 6
Case-II If C is right on A
⇒ D must be right on B
= (4 - 1)! = 3! = 6
Hence total no. of ways is 6 + 6 + 6 = 18
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