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.
Best Answer
Let us approach part (c) constructively. You appear to have correct answers for all the parts, but perhaps this will help.
First we select two of the four floors, which can obviously be done in $\dbinom{4}{2} = 6$ ways. Now we select two people from the four people to leave on the first floor, which can be done in $\dbinom{4}{2} = 6$ ways. We see that there are a total of $6 \times 6 = 36$ successful outcomes.
There are a total of $4^{4} = 256$ ways. Our probability is thus $\frac{36}{256} = \boxed{\frac{9}{64}}.$