The question is explicitly: "In how many different orders can $n$ cars line up in a gas station of 4 parallel gas pumps?" No more information is provided. I initially assumed that both the pumps and cars are distinguishable.
In these cases, different orders count towards the total number of possible outcomes, and repetition is not allowed since the same car can't be parked in two locations. Would that mean the solution would take the form $P(n,r)$? But that would reduce to $n!$ which is just the number of different arrangements the cars can take in a single line.
Would I just multiply by $4$ to get the number of arrangements if they're distributed across $4$ gas stations?
Instinctively it should look more like
(n - # of cars @ 2,3,4)! + (n - # of cars @ 1,2,4)! + (n - # of cars at 1,2,4)! + (n-# of cars at 1,2,3)!
How can I express this purely in terms of n?
Method 1: We split the $n$ cars into four possibly empty groups, then multiply by the number of ways of lining them up.
The number of ways the cars can be split into four groups is the number of solutions of the equation $$n_1 + n_2 + n_3 + n_4 = n$$ in the nonnegative integers. A particular solution corresponds to the placement of three addition signs in a row of $n$ ones. For instance, if $n = 10$, then $$1 1 + 1 1 1 + + 1 1 1 1 1$$ corresponds to the solution $n_1 = 2$, $n_2 = 3$, $n_3 = 0$, and $n_4 = 5$. The number of solutions of the equation $n_1 + n_2 + n_3 + n_4 = n$ in the nonnegative integers is $$\binom{n + 3}{3}$$ since we must choose which $3$ of the $n + 3$ locations required for $n$ ones and $3$ addition signs will be filled with addition signs. The cars can be lined up in $n!$ orders. Once an order is chosen, we send the first $n_1$ cars to pump 1, the next $n_2$ cars to pump 2, the next $n_3$ cars to pump 3, and the remaining $n_4$ cars to pump 4. Hence, the number of distinguishable ways the $n$ cars can be distributed to the four pumps is $$\binom{n + 3}{3}n!$$
Method 2: The $n$ cars can be lined up in $n!$ ways. However, we also have to choose $3$ points where we divide the line. Hence, we have to arrange $n$ distinguishable cars and $3$ indistinguishable dividers. Hence, we have $n + 3$ objects to arrange. We choose $3$ of those $n + 3$ positions for the dividers, then arrange the $n$ cars in the remaining $n$ positions in $n!$ orders. Therefore, there are $$\binom{n + 3}{3}n!$$ ways to arrange the $n$ cars at the four gas pumps.