A roller coaster has five cars, each containing four seats, two in front and two in back. There are 20 people ready for a ride. In how many ways can the ride begin? What if a certain two people want to sit in different cars?
Every person can pick a seat. So there is 20! seat arrangements. Something is wrong with this reasoning, what is it about five cars which makes me miscount the number of ways the ride can begin?
If, instead of a roller coaster there was a ferris wheel one wouldn't be able to apply the "pick a seat" logic because it's the wheel that is round, not the seats(not $19!$) and a natural choice would be to put people in groups of 4 and then permute the groups on the wheel $5!/5$ and then permute the people in the cars, ${4!}^{5}$.
Which finally gives us $20!4!$ ways the ride can begin.
Now, I'm wondering, if the same approach was taken(split into groups) for the roller coaster there would be $20!5!$ ways to start a ride.
Equivalent problem would be -> There's a giant concert hall with 100 seats. There are 50 blue and 50 white seats. In how many ways can the concert begin if 100 people want to listen to it?
I would say 100! but since colors can be interpreted like cars this would mean that splitting people in 2 groups of 50, picking a car of colored seats would take 2! the final solution with the ferris wheel approach would give me $100!2!$.
I have no idea which of the 100! ways I lost in the first solution.
What I'm asking is what and where is the flaw and then what is the generalized solution?