Number of ways to visit each of $8$ zoos at least once and at most $3$ times Studying for the midterm. Trying to solve the following problem:

There are $8$ zoos in the City. During the next $13$ weeks, each week Benny wants to visit one zoo. Count the number of possibilities so Benny visited each zoo at least once and at most three times.

What I did: I want to fulfill the "at least one" request. So let's choose $8$ weeks in ${13 \choose 8}$ possibilities. Then I arrange the $8$ zoos in $8!$ possibilities. Now I left with $5$ additional weeks that should be filled with the zoos. But how can I fulfill the "at most three times"?
 A: @Math Lover gave you all possible ways. I want to solve this question using $2$ approach ,i.e generating functions.
Let assume that we align all $13$ weeks in a row such that $$-,-,-,-,-,-,-,-,-,-,-,-,-$$
The first line means the first week , the second line means second week so on.
We want to put these $8$ distinct zoos to these lines with obeying the given restriction.For example , one of the possible visiting order is : $$1,3,5,5,7,8,6,1,1,3,4,4,4$$ where each number represent one of these $8$ zoos. This arrangements tells us  we visit first zoo in first week , visit third zoo in second week etc.We could also arrange them like $$3,1,5,5,6,8,7,1,3,1,4,4,4$$
As you realize we are making permutation with repetition like the questions asking "how many possible words are there using the letters of MISSISSIPI". However , we have an hindrance here such that the number of visiting zoos range from $1$ to $3$.
Then , exponential generating functions comes to help !
For given restrictions E.G.F. of each zoo is equal to: $$\bigg(x + \frac{x^2}{2!}+ \frac{x^3}{3!}\bigg)$$
Realize that $x$ means given zoo is visited once , $x^2$ means given zoo is visited twice , $x^3$ means given zoo is visited three times.
Then , we should find the coefficient of $x^{13}$ and multiply it by $13!$ or find the coefficient of $\frac{x^{13}}{13!}$ in the expansion of $$\bigg(x + \frac{x^2}{2!}+ \frac{x^3}{3!}\bigg)^{8}$$
So $$13! \times \frac{119}{12} =61.751.289.600$$ ways there are
