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.

Your calculation for $c)$ is correct, though, as the comments and answers show, you seem not to have expressed it in a readily comprehensible manner.
I would express what I understand your argument to be as: There are $3$ ways to split the $4$ people into $2$ groups. Then we can treat those two groups like $2$ people and choose the floors they get off at, $4$ options for the first group and $3$ for the second, or $12$ overall, for a total of $3\cdot12=36$ options out of $4^4=256$.
