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.

Solutions using conditional probabilities:
Let's enumerate people $P_1, P_2, P_3, P_4$.
a) all exit at different floors
$P_1$ can pick any of 4 floors with probability $\frac{1}{4}$ each, so we have 4 cases. Because all 4 cases are equivalent we focus on just one. Now, given the floor $P_1$ exited at, $P_2$ can exit on any of the 3 floors left with probability $\frac{3}{4}$, then $P_3$ can exit on any of the 2 floors left with probability $\frac{2}{4}$, and finally, $P_4$ only one floor left with probability $\frac{1}{4}$. Thus, multiplying these conditional probabilities together:
$\sum_{floor=1,4}{\frac{1}{4} * \frac{3}{4} * \frac{2}{4}} * \frac{1}{4} = \frac{3}{4} * \frac{2}{4} * \frac{1}{4} = \frac{6}{64} = \boxed{\frac{3}{32}}$
b) all exit at the same floor
Again, $P_1$ can pick any of 4 floors with probability $\frac{1}{4}$ each, and then the rest of people have a probability $\frac{1}{4}$ to match that floor:
$\sum_{floor=1,4}{\frac{1}{4} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4}} = (\frac{1}{4})^3 = = \boxed{\frac{1}{64}}$
c) two get off at one floor and two get off at another
Given $P_1$ picked a floor, we have 4 cases with probability $\frac{1}{4}$ each: $P_2$ picks the same floor with probability $\frac{1}{4}$, and the other two a different floor with probability ($\frac{3}{4})^2$:
$\sum_{floor=1,4}{\frac{1}{4} * (\frac{3}{4})^2} = \frac{1}{4} * (\frac{3}{4})^2 = = \boxed{\frac{9}{64}}$
Now, logical question would be what if it's not $P_2$ that exits on the same floor with $P_1$ but rather $P_3$ or $P_4$? There are $3!$ combinations that may happen with equal probabilities of $\frac{1}{3!}$, which would eliminate each other. But, to be complete here is the full formula:
$\sum_{floor=1,4}{\frac{1}{4} * \sum_{cases=1,3!}{\frac{1}{3!} * \frac{1}{4} * (\frac{3}{4})^2}} = \frac{1}{4} * (\frac{3}{4})^2 = = \boxed{\frac{9}{64}}$
The same reduction, of course, applied to a) and b).
