What is the probability that each inhabitant of a three-story building lives on a different floor? 
There is a multi-apartment building with $3$ stories and $4$ apartments at each story. In each apartment lives one person. Three random inhabitants of this building are standing outside the building. What the probability that each of them live on a separate floor (event B).

I want to solve this problem using combinatorics approach. The answer that I have in my book is: 
$$P(B) = \frac{|B|}{|\Omega|} =  \frac{12 \cdot 8 \cdot 4}{12 \cdot 11 \cdot 10}$$
To best of my understanding the logic goes as follows:
(a) total number of possibilities:  ordered sample ($3$ out of $12$)
(b) needed possibilities: first we take a person from any apartment ($12$ possibilities), then a person from two remaining floors ($8$ possibilities) and  finally a person from one remaining floor ($4$ possibilities).
My question: When I choose people in this problem I do not think we care about the order. So, I think the sample should be unordered. Probably mathematically it does not matter because the “order” factor is in both numerator and denominator is the same. But if indeed I solve the problem as the unordered sample, how I calculate the possibilities in numerator ($|B|$)?
Many thanks.

You can indeed solve this problem by counting (unordered) combinations of people. In this case the numerator would be the number of ways to choose $3$ people out of the $12$ such that exactly one person lives on each floor. Since $4$ people live on each floor, that would be:
$$4*4*4=64$$
