Elevator Probability Question There are four people in an elevator, four floors in the building, and each person exits at random. Find the probability that:
a) all exit at different floors 
b) all exit at the same floor 
c) two get off at one floor and two get off at another 
For a) I found $4!$ ways for the passengers to get off at different floors, so $$\frac{4!}{4^4} \text{would be the probability} = \frac{3}{32}$$ 
For b) there are only four ways for them to all exit on the same floor, so $$\frac{4}{256} = \frac{1}{64}$$ 
For c) am I allowed to group the $4$ people so that I am solving for $2$ people technically? For two people there would be $12$ possibilities, and there are three ways to group the $4$ individuals, so $$\frac{12 \cdot 3}{256} = \frac{9}{64}$$ 
I'm not sure if I'm doing these right, can you please check? Thank you.

a and b are correct.  For c, I would say there are ${4 \choose 2}$ ways to choose the first pair, $4$ ways to choose the floor they get off on, and $3$ ways to choose the floor the other pair gets off on, but we need to divide by $2$ because we can swap the pairs.  That gives $\frac {6\cdot 4 \cdot 3}{2 \cdot 4^4}=\frac 9{64}$  This is the same as your answer, but I don't understand the logic that gives $12$ possibilities for a pair.  We don't care about the order of selection in a pair.
