Why is this considered ordered sampling (Casella Berger)? I'm working on example 1.18 in Casella & Berger and I'm confused by the solution.
The problem states:

If n balls are placed at random into n cells, find the probability that exactly one cell remains empty.

In the solutions manual, it states that there are $n^n$ total ways to place $n$ balls into $n$ cells. This implies the sampling method is ordered with replacement. I agree that the $n$ cells are sampled with replacement. But, why is it considered ordered?
My logic:
Let's pretend there are 3 cells. If I place a ball into {1,1,2}, {1,2,1}, or {2,1,1}, in the end I have 2 balls in cell 1, and 1 ball in cell 2. The ordering does not matter. 
What am I misunderstanding here?
 A: Since this problem is self study (and the solution is given in the text) I'll sketch out how I would approach the problem and let you fill in the gaps. I think it's easiest to approach by counting the valid orderings (like Michael suggested in the comments) and dividing by the total possible orderings. 
The problem becomes easier when you think about what must be true if only one cell is empty. Here's a hint: if precisely one cell is empty, can a cell have 3 balls in it? 
Solution hidden below:

 No. If precisely 1 cell is empty, then precisely 1 other cell has 2 balls in it and the rest have 1 ball. If one cell had 3 balls then at least 2 cells must be empty. 

Given that, it's straightforward to count the number of valid events and total events. 
How many ways can we choose the 2 cells which don't have 1 ball in them? 

 ${n}\choose{2}$

How many ways can we order the balls and place them into cells?

 $n!$

That gives us our numerator, the total number of possible orderings: 

 ${{n}\choose{2}} \cdot n!$

How many ways can I place $n$ balls into $n$ cells? Well the first ball can be placed into $n$ cells, the second into $n$ cells, and so on leading to $n^n$ total possible arrangements of the system. This gives the probability as (number valid)/(total): 

 ${{n}\choose{2}}\ n!\ /\ n^n$

