Question 1036511: Hi , im really not sure about how to go about and solve this question, i tried it and i got 3C3 * 3! *6! where 3C3 is the 3 blanks filling in with 3 boys , 6! ways to seat the girls and 3! due to order importance, could you please help me with the question
1) In how many ways can 6 girls and 3 boys sit on a row of 9 chairs in such a way that no two boys sit next to each other?
Thanks
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! There are 6! ways of seating the 6 girls. Then there are 7 spaces in which you can insert the boys (including both ends), and this can be done in 7P3 ways. Therefore by the fundamental principle of multiplication, there are 6!*7P3 = 151,200 ways in which no two boys sit next to each other.
|
|
|
