Why is this a permutation instead of a combination? Textbook question: (apologies for the "baby math" question)

A new company with just two employees, Sanchez and Patel, rents a floor of a building with 12 offices. How many ways are there to assign different offices to these two employees?

Textbook answer is $12 \cdot 11 = 132$ by the product rule. Ok cool. Now this was before it introduced permutations and combinations, but I can see it's a permutation $P(12, 2) = \frac{12!}{10!} = 132$.
My question is, why is this a permutation and not a combination? Permutations only apply when the order of the resulting subsets is involved, so how does the order of the office assignments matter in this problem? I get mechanically that it uses the product rule (i.e. pick from 12 then pick from 11) but what I don't understand is, since this is the same value as the permutation then apparently the order matters, or they wouldn't come out to the same answer. So how does order play into this?
 A: It is a permutation because assigning Sanchez to office 1 and Patel to office 2 is different from assigning Patel to office 1 and Sanchez to office 2.  
A: You can see this problem as an one-to-one function where both objects(employees) and images(offices) are labeled/distinct. The function must be one-to-one because each office can only be assign to one employee.
There is a way to solve all the combinatoric kind of problems. It's the Twelvefold Way. You can google it.
A: A simple way to understand:
a permutation is nothing but [choose]$\times$[place] i.e. $ ^nP_k =\; ^nC_k\times k!$
So we choose $2$ offices out of $12$, and place $S$ and $P$ in them in $2!$ ways.
A: Let the $12$ offices be numbered $1$ to $12$. One way in which different offices can be assigned to Sanchez and Patel is to assign office $7$ to Sanchez and office $3$ to Patel, denoted by $(7,3)$.  Another assignment is $(2,12)$.  The number of ways to fill in the two blanks in $(\_,\_)$ is the number of ways to fill the first blank (which is $12$) times the number of ways to fill the second blank (which is $11$ because the second blank can be any office except the one already assigned in the first blank).  So, the number of ways in question is $12 \times 11$.  Note that this is a permutation problem because the order matters - in this problem, the assignment $(7,3)$ is different from the assignment $(3,7)$.   
One could think of a problem where you want to form a team of two players from a total of $12$ available players.  The number of possible teams you can form is ${12 \choose 2} = 12 \times 11 / 2$.  We are dividing $12 \times 11$ by $2$ here. This is a combinations problem because the order doesn't matter - the two subsets $\{7,3\}$ and $\{3,7\}$ refer to the same team.  
