+4 This was the answer I found online:
If the twins sit at the end of the row, then the other sister can sit at one of 5 places so that she’s not next to the twins. 1st choose one of the 2 ends (2 choices) and seat the 2 twins (2! choices); then seat the other sister (5 choices); then seat the remaining 5 brothers (5! choices).
2! X 2! X 5! X 5! = 2400 ways
But it doesn't mention in the problem that the boys have to be seated always adjacent to each other...so that means I'm assuming you can have a configuration like this:
BTBGBBB
T representing the twin girls..
My calculation was as follows...
2! * 7 ways to place the twins in the 8 seats.
in (2!*5) of those configurations you have 4 places to place the other girl, so (2!*5)*4 = 40.
in (2!*2) - the ends of the row - we have 5 places to place the other girl, so (2!*2)*5 = 20.
5! ways to place the boys in any of these configs, so (40+20)*5! = 7200.
Is this correct or incorrect?
