Re: Circular Permutations

Quote:

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!

Your method is correct but your counting is off.

Recall that $\displaystyle \|A\cup B\|=\|A\|+\| B\|-\|A\cap B\|$, where $\displaystyle \|A\|$ is the number of cases in $\displaystyle A$.

You found the first two but not the case in which Jane is seated between Ben & Roy. i.e. BJR or RJB.

Re: Circular Permutations

Quote:

Originally Posted by

**Plato** Your method is correct but your counting is off.

Recall that $\displaystyle \|A\cup B\|=\|A\|+\| B\|-\|A\cap B\|$, where $\displaystyle \|A\|$ is the number of cases in $\displaystyle A$.

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?

Re: Circular Permutations

Quote:

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.