# Is There a Better Strategy for this Combination Scenario?

An elevator containing five people can stop at any of seven floors.
What's the probability that no two people get off at the same floor?
Assume that the occupants act independently
that all floors are equally likely for each occupant.

For the denominator, since each person has seven choices,
there are $7^5$ way that the passengers can occupy the 7 floors.
Then, for the numerator, let the 5 passengers form one group,
and 2 imaginary passengers forming a second group.
So there are 7! permutations, but since the order among the 5 passengers and the order among the 2 imaginary passengers do not matter, you have $7!\over 5!2!$ ways that the 5 passengers can occupy the floors such that each floor has only one or zero passenger. Altogether, you have $7 \choose 5$ / $7^5$.

I am pretty sure that I got the numerator correct.
It took me a long time to figure out the appropriate analogy;
is there a more direct method to this question?
Furthermore, why is the denominator $7^5$ and not $6^7$?
In the context of this question, $6^7$ means that each floor can take 0, 1, ... 5 passengers.

Thanks!

Note that in your denominator, you've allowed different arrangements of people on the same floors to count multiples times i.e. Person A on Floor 1 and Person B on Floor 2 is counted differently to Person A on Floor 2 and Person B on Floor 1 hence your numerator should also allow different arrangements to be counted. So the numerator should be $^7P_2=\frac{7!}{2!}$ not $^7C_2$.
The denominator wouldn't be $6^7$ as you claim because you've removed all dependency on the number of people. This method of counting would allow for anywhere between 0 and 35 people.
You are correct that each floor can have $0$ to $5$ occupants, but they are not independent. If one floor has $5$ all the others must have $0$. $7^5$ is correct for the denominator.