Q) Six persons A , B , C , D , E and F are to be seated around a circular table. Find the number of ways in which 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) Consider three cases:

One case with A, having B to his right, and C to B's right.

Now number of ways the rest can be arranged is 3!.

For the next case, take A, then B to his right, and D to B's right... again the rest can be arranged in 3! ways.

For the last case, you have A, with C to his right. Now three entities are left - E, F and BD (since B must be seated with D to his immediate right 'cos C is already used up) and again we get 3! ways.

So total number of ways is 3. 3! = 18 ways

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