1. ## Circular Permutations

Hi guys, need with with this question:

In how many ways can 5 gentlemen, involving Roy and Ben, and 5 ladies, including Jane, be seated at a round table, such that - Jane sits next to neither Roy nor Ben?

This is how I did it:
Case 1 - Jane sits next to Roy
Grouping Jane and Roy, no.of ways = (9!/9) * 2! = 80640

Case 2 - Jane sits next to Ben
No.of ways = (9!/9) * 2! = 80640

Taking complement,
(10!/10) - 2(80640) = 201600

Is my method wrong? Thanks in advance!

2. ## Re: Circular Permutations

Originally Posted by Blizzardy
In how many ways can 5 gentlemen, involving Roy and Ben, and 5 ladies, including Jane, be seated at a round table, such that - Jane sits next to neither Roy nor Ben?

This is how I did it:
Case 1 - Jane sits next to Roy
Grouping Jane and Roy, no.of ways = (9!/9) * 2! = 80640

Case 2 - Jane sits next to Ben
No.of ways = (9!/9) * 2! = 80640

Taking complement,
(10!/10) - 2(80640) = 201600

Is my method wrong? Thanks in advance!
Recall that $\|A\cup B\|=\|A\|+\| B\|-\|A\cap B\|$, where $\|A\|$ is the number of cases in $A$.
You found the first two but not the case in which Jane is seated between Ben & Roy. i.e. BJR or RJB.

3. ## Re: Circular Permutations

Originally Posted by Plato
Recall that $\|A\cup B\|=\|A\|+\| B\|-\|A\cap B\|$, where $\|A\|$ is the number of cases in $A$.
You found the first two but not the case in which Jane is seated between Ben & Roy. i.e. BJR or RJB.
Thanks! But to calculate case 3, do I just group BJR together?
i.e. No. of ways (Case 3) = (8!/8) * 2! = 10080 (Only arrange B and R)
So, taking complement, (10!/10) - 2(80640) - 10080 = 191520
Is this correct?

4. ## Re: Circular Permutations

Originally Posted by Blizzardy
Thanks! But to calculate case 3, do I just group BJR together? i.e. No. of ways (Case 3) = (8!/8) * 2! = 10080 (Only arrange B and R) So, taking complement, (10!/10) - 2(80640) - 10080 = 191520
I have not done the actual calculations. But the method is correct.