In how many ways can we arrange $80$ people in $5$ cars, so that in the first car there are exactly $15$ people? Also, in how many ways can we arrange them if $15$ people should be in one of the cars?
I used ‘stars and bars’ method.
Let $x_i$ be number of people in car i;
So $x_1+x_2+x_3+x_4+x_5=80$
In the first question $x_1=15$, so $x_2+x_3+x_4+x_5=65$, that's $\frac{(65+3)!}{(65! \times 3!)}$
But I am stuck on question $2$. Because if we multiply $\frac{(65+3)!}{(65! \times 3!)}$ by $5$, we will get more arrangements, from what are possible because we count multiple times those arrangements when in at least $2$ cars there are exactly $15$ people.
 A: With your assertion that only numbers matter for people, and cars are distinct, the first part is ok.
For the second part, apply stars and bars with inclusion-exclusion. 
(15 people in at least one car - 15 people in at least two cars + ....)
$=\dbinom51\dbinom{68}3 - \dbinom52\dbinom{53}2 + \dbinom53\dbinom{38}1 - \dbinom54\dbinom{23}0$
A: Of course the fact that the problem is about people rather than red balls or cases of beer suggests that you should consider that the people are distinguishable. 
The question remains if the cars are distinguishable. I will assume that they are, and will also assume that the car that should have 15 people has already been chosen.
You start by picking 15 people for the first car:
$$
{80\choose15}=\frac{80!}{15!(80-15)!}=6635869816740560.
$$
Then you need the number of ways to fill exactly 4 cars with the remaining 65 people. Note that this is the number of ways to fill at most 4 cars minus the number of ways to fill at most 3 cars, i.e.
$$
4^{65}-3^{65}.
$$
Therefore the number is
$$
{80\choose15}\times(4^{65}-3^{65})=9032277882922331201019483936739005898712103023525609360
$$
or $\approx9\times10^{54}$.
