Counting: A distributed network of 10 servers, 40 different movies will be stored on network 
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*How many ways are there to select where the movies will be stored if there is no restriction on the number of movies stored at each server?
Answer: $10$^$40$
Comparing this problem to problems like how many different passwords are there of length $6$ using letters only. The letters are constant and never need refilling, and you always have $26$ possibilities per slot => (26^6) total outcomes. However I thought that the $40$ movies are finite, and storing some would decrease the overall count each time you store them, something like a permutation/factorial. Why can I store the movies as if they refill/are constant?
2. How many ways are there to select where the movies will be stored if the same number of movies are stored at each server?
Answer: $40! / (4!)$^$10$ 
I am lost as to why factorials were used and what they represent. Now treating the movies as a finite resource compared to problem #1. Also confusing where the $4!$ in the denominator came from. Can someone explain the reasoning?
 A: For the first question, what is 'refilling' is associated with the base of the exponential, in $26^6$, 26 is the number of leters, and in $10^{40}$, 10 is the number of servers (which are infinite in space as you said), it's not the number of movies (which you want to store in a single server each).
For the second one:
Think about how many options you have at each step of making such a distribution.
If all the servers have the same number of movies, then all of them have 4 movies each.
Let's choose the movies one server at the time: 
-For the first one you have to choose 4 of the 40 movies and like that you have $\dfrac{40!}{4!(40-4)!}$ options (do you see why?).
-When choosing the movies of the Nth server, you have already distributed $4(N-1)$ movies, so there are $40-4(N-1)$ left to choose from. For this server we have to choose 4 movies, so we have $$\dfrac{(40-4(N-1))!}{4!(40-4(N-1)-4)!}=\dfrac{(40-4(N-1))!}{4!(40-4N)!}$$
Now for the total amount of choices we have to do the product of the amount of choices at each step. But notice that $(40-4N)!$ is in the denominator for the Nth step and is in the numerator for the (N+1)th step, so they cancel out when making the product.
The only occurrence of this type that remains is the last one, because there is no 'next' term for cancelling. But in the last occurrence, $N=10$, so it's just (40-40)!=1.
So the total product is $\dfrac{40!}{4!^{10}}$.
