A difference between a roller coaster and a ferris wheel. 
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?
 A: The mistake that you made with the Ferris wheel is as follows:
You started off picking the groups of 4 people as ${ 20 \choose 4} \times { 16 \choose 4} \times { 12 \choose 4} \times {8 \choose 4} \times { 4 \choose 4}$. However, choosing ABCD first and then EFGH, is the same as choosing EFGH first and then ABCD. As such, you over counted $5!$ times, and hence need to account for that. This is where the 'extra' 5! comes from in your 'equivalent roller coaster'.
The simple way to see why this is wrong, is to think of how many ways there are to choose 5 groups of 1 people from a group of 5. Clearly, there is only 1 way to do so. However, if we were to use your explanation, there are 5 ways to pick the first person, 4 ways to pick the second, 3 to pick the third, 2 to pick the fourth, 1 to pick the fifth. What you are doing, is permuting the 5 groups, as opposing to choosing the 5 groups.

This should be enough information for you to fix your 'concert hall' issue. Being confused between permuting and choosing of groups is an extremely common situation. It occurs because you do not consciously realize that you're actually introducing order into that groups.
A: In the spirit of Gerry Myerson's comment, you need to define what makes one configuration different from another.  If the two people in the front seats of a car switch places, is that different?  Are the cars numbered so if everybody moves one car aft you can tell the difference?  If the answer to both of these is yes, then $20!$ is correct.  If not, you need to divide by the number of times that each configuration is counted.  So if you don't care if a rider is on the left or right in a car, there are $10$ pairs that can switch so you divide by $2^{10}$.  If the cars are interchangeable, you can rotate $5$ ways, so you divide by $5$.  Without a clear question, there is no clear answer.
