An elevator containing five people can stop at any of seven floors. An elevator containing five people can stop at any of seven floors. What is the probability that no two people get off at the same floor? Assume that the occupants act independently and that all floors are equally likely for each occupant.
Solution:  I don't really have too much issue with the solution except that I keep on over counting and I'm curious as to what is it extra that I am considering that I should not be considering?
My solution:  $$P(A) = \frac{(7)(6)(5)(4)(3) 5!}{7^5}$$
Now I am aware that the $5!$ should not be there, but my reason for including it is because we do not know which of the 5 people is the "first" person, "second" person, etc...
 A: This is easiest if we treat the people as being distinguishable.
There are $7$ floors each person can go to, and since the floors are chosen independently, there are $7^5$ total ways for the people to get off of the elevator.
There are $\binom{7}{5} = 21$ ways to choose different floors for all $5$ people, and having chosen these floors, there are $5!$ ways to permute the people among them. Therefore, of the $7^5$ total arrangements of people, $\binom{7}{5}5!$ have no two people on the same floor. The probability of this happening is
$$\frac{\binom{7}{5}5!}{7^5}.$$
This is only slightly harder if you assume the people are identical, but you will then end up with a probability of $$\frac{\binom{7}{5}}{\binom{11}{6}}.$$ I think that the first interpretation is probably what was intended.
A: Calling the persons $P_i$ there is a probability of $7^{-5}$ that person $P_i$ steps out on floor $i$ for every $i\in\{1,2,3,4,5\}$. 
There are $5!$ rearrangements for the persons that result in a step out of exactly one person at the floors $1,2,3,4,5$. 
Moreover there are $\binom75$ ways to choose $5$ floors. 
So we end up with probability $$5!\binom757^{-5}$$ 
