Seven married couples, five of which with a single child, go to a movie theater. In how many ways can they line up to buy tickets so that no two children stand next to each other?
This is the problem, word for word.
The problem does not state that families have to be together. The part (b), which I have done already, did specifically say families are together. So, this is being done on the assumption that since it was not stated families are together, they do NOT have to be together.
Arranging them without restriction = 19!
choosing combos: n!/(r!)(n-r)!
The Attempt at a Solution
Here is my thought process:
10 slots in the line have to be filled by a pair of child/adult. So I did 19!/(10!)(9!). That gives the number of different ways that those slots can be filled. Then there are 9 remaining slots that are filled by 9 adults, so I did 9!. So the whole thing would be:
19!/(10!)(9!) * 9! = 19!/10!
I just have a feeling that that's not right.