9 people drive into a parking lot, 4 are boys and 5 are girls. They decide to park in a straight row of 9, if none of the girls park next to each other, how many different parking arrangements are there?
Thanks
Hello, RicLang!
9 people drive into a parking lot, 4 are boys and 5 are girls.
They decide to park in a straight row of 9 spaces.
If none of the girls park next to each other, how many different parking arrangements are there?
If the girls are non-adjacent, there is ONE arrangement of the genders: .$\displaystyle GBGBGBGBG$
Then the five girls can be arranged in: .$\displaystyle 5!$ ways.
The four boys can be arranged in: .$\displaystyle 4!$ ways.
Therefore, there are: .$\displaystyle (5!)(4!) \:=\:(120)(24) \:=\:2880\text{ arrangements.}$