A confusing counting problem The problem says:

$12$ persons have at their disposal the $3$ cars of $6$, $4$, and $2$ seats respectively.
  How many ways can we assign these $12$ people to the three cars supposed:
  
  
*
  
*Anyone can drive?
  
*Only $4$ of $12$ can drive?
  



*

*The answer of the first question is easy:
$${12 \choose 6}{6 \choose 4}{2 \choose 2} = 13860$$

*In this part, I solved as follows:


$$ 3! {4 \choose 3}{9 \choose 5}{4 \choose 3}{1 \choose 1} = 12096 $$
However, my teacher solved as follows: 
$$   {4 \choose 3}{9 \choose 5}{4 \choose 3}{1 \choose 1} = 2016 $$
i.e. he didn't respect the order present in the drivers' seats. He considered that all the drivers' seats are alike which is not true(in my opinion)
My questions are:


*

*Am I, my teacher, or neither of us right?

*Concerning both answers of the $2$nd part of the problem: I faced another issue that if I wanted to solve the first part again using the same way of thinking in the $2$nd part I get a completely different answer.
That is if $12$ persons can drive, why can't we say:
Number of ways = 
$$ 3! {12 \choose 3}{9 \choose 5}{4 \choose 3}{1 \choose 1} = 665280 $$
 A: The important point here is that you only care about which car each person is in. You don't care whether they are driving car I, you just care whether they are in car I.
This should answer the alternate (wrong) method which you used to solve the 1st question - you cared about who is driving each of the cars when you shouldn't. (For example, if A and B are in the two-seater, it doesn't matter whether A is driving or B is driving.)
Now, for the second question, you want to ensure that each of the three cars has at least one driver. As you wrote, the number of ways should be $3! {4 \choose 3}{9 \choose 5}{4 \choose 3}{1 \choose 1}$. However, as the car with 2 drivers can have two possible people driving, we must divide by a $2$ and the answer should be $3 {4 \choose 3}{9 \choose 5}{4 \choose 3}{1 \choose 1}=6048$.
A: These are two different ways of counting this. You can either say: assign persons "1,4,5" to car "1" , or assign persons "1,4" to the backseat of car "1", person "5" to the driver-seat. The answer depends on whether you consider picking a designated driver as part of the assignment. With driver selection, you would get 
$$3! {12 \choose 3}{9 \choose 5}{4 \choose 3}{1 \choose 1} = 665280$$ in the first part and $$3! {4 \choose 3}{9 \choose 5}{4 \choose 3}{1 \choose 1} = 12096$$ in the second part. Let's say that driver selection is not part of the assignment. Then we simply have to have people who can drive inside each car, but no designated driver. We can obtain the solutions by "forgetting" the designated driver of our previous method:
$$\frac{665280}{6*4*2} = 13860$$ in the first case.
In the second case, having 4 drivers for 3 cars, each car can have only 1 possible designated driver, except the car with 2, leading to:
$$\frac{12096}{2} = 6048$$ 
