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(20054)   (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
|
|
|
