# of seating arrangement in a 6 seat car I'm getting hung up on a probability question:
A car has six seats including the driver’s, which must be occupied by a driver. In how many ways is it possible to seat 4 people if all 4 can drive.
for one possible way you can choose the driver first and then choose seat options per person left so that would look like:
$$4*5*4*3$$
what I keep getting hung up on is when I try to determine seat options
so the first seat has 4 options since all 4 people can drive and it must be chosen.
seat 2: I think should be 4, 3 people can choose it or it can be left empty
seat 3: should be 3, 3 people can choose it if the seat 2 isn't chosen, or 2 people can choose it if seat 2 is chosen, or it can be left empty
seat 4: 5 possibilities; if seat 2 isn't chosen and seat 3 isn't chosen then 3 people can choose this, or if seat 2 is chosen and seat 3 isn't chosen then 2 people can choose it, or if seat 2 and 3 are chosen then 1 person can choose this seat, or if seat 2 isn't chosen and seat 3 is chosen then 2 can choose it, or it can be left empty
seat 5:  no clue
seat 6: no clue
this is a lot of text and should be easier (I think) but this is the process my mind is taking.
any help would be greatly appreciated.
 A: Your first way of counting is perfectly good. 
If you want to count another way, let us invent $2$ identical ghosts. The seats for them can be chosen in $\binom{5}{2}$ ways, since ghosts aren't allowed to drive. (There  is a problem with taking their picture for the licence.) The rest of the seats can be filled in $4!$ ways.  
A: There are three parts to this.  There's always a driver and all 4 can drive so whatever the combination in the other 5 seats, you are going to multiply that combination by 4 to represent the 4 different sets of people who could be in the other seats.
Next, you have three people to fill five seats.  You will want to determine the number of combinations of filled seats and the order of the 3 people as separate factors.  If seats 1 & 2 are always occupied, you have 3 combinations.  If seats 1 & 3 are always occupied, you have 2 new additional combinations. If seats 1 & 4 are occupied, there is 1 new combination.  So if seat 1 is always occupied, there are 6 combinations.  Now, if seat 1 is always vacant, you are distributing through seats 2 through 5.  Same method of counting shows 3 new combinations with seat 1 vacant and seat 2 occupied. Finally, if you vacate seats 1 & 2, you have one new combination.  So, that's 10 combinations of occupied seats.  Then there is the order of each person sitting for any particular combination of occupied seats. That's a simple 3x2. So 4 (driver sets) x 10 (seating combinations) x 6 (order of people) = 240 combinations.
Now, we want more mathematical eloquence than just counting the five seat positions.  If we were dealing with a genome instead of a 6 passenger car, counting wouldn't be very practical.  To count the combination of occupied seats, I did that counting occupied seats.  I could have counted vacant combinations. If seat 1 is vacant there are 4 combinations for the other vacant seat. If seat 2 is always vacant there are 3 new combinations, seat 3 vacant adds 2 new combinations and seat 4 one new combination.  That still adds up to 10, but it is still not eloquent.  If we are using Google sheets, we could use combin(5,2) to get the number of combinations, so what is the math behind combin?  See, https://corporatefinanceinstitute.com/resources/excel/functions/combin-function/ for the actual formula.  
Please point out my errors and be critical of the answer.  I'm old and have forgotten a lot.
