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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!
Your method is correct but your counting is off.Quote:
Originally Posted by Blizzardy
Recall that, where
is the number of cases in
.
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?Quote:
Originally Posted by PlatoYour method is correct but your counting is off.
Recall that
You found the first two but not the case in which Jane is seated between Ben & Roy. i.e. BJR or RJB.
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?
I have not done the actual calculations. But the method is correct.Quote:
Originally Posted by Blizzardy