SOLUTION: A group of 7 children is going to a movie. In How many ways can the group be seated, if Randy, Sandy, Candy and Dandy refuse to sit next to each other?

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Question 890592: A group of 7 children is going to a movie. In How many ways can the group be seated, if Randy, Sandy, Candy and Dandy refuse to sit next to each other?
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
A group of 7 children is going to a movie. In How many ways can the group be seated, if Randy, Sandy, Candy and Dandy refuse to sit next to each other?
Let the seat numbers be:  1,2,3,4,5,6,7

Randy, Sandy, Candy and Dandy must sit in the 4 odd numbered seats.
They can be placed in the 4 odd numbered seats in 4! ways.
The other three children must be placed in the 3 even-numbered seats.
They can be placed in the 3 even numbered seats in 3! ways. 

Answer 4!*3! = 24*6 = 144 ways.

Edwin

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