Four Person Elevator

There are four people on the ground floor of a building that has five levels not including the ground floor. They all get into the same elevator.

If each person is equally likely to get on any floor and they leave independently of each other, what is the probability that no two passengers will get off at the same floor?
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To start out with we need to understand exactly what the probability is that we’re calculating. If “no two passengers can get off on the same floor”, then that essentially means each person has to get off on their own floor. 

We then need to get the total number of combinations for each person and each floor. That way we can get the denominator for our fraction when we’re considering the possibility of two people not getting off on the same floor. 

The number of ways to assigning five floors to four different people is to get the total sample space. In this case it would be 5 * 5 * 5 * 5. For each person, they can choose one of five floors, which happens four times for four people. So the total number of combinations is:

5
4
=
625
5 
4
 =625

Now how do we compute the number of combinations of users without repetition of floors? Well for the first passenger you have five different options. The second person has four, and so on. Note that this number counts all possible orders betwen passengers as well. 

The number of ways to assign five floors to four people without repetition of floors is then each passengers total combinations of 5 * 4 * 3 * 2.

The result is then
5
5
⋅
4
5
⋅
3
5
⋅
2
5
=
0.192
5
5
​
 ⋅ 
5
4
​
 ⋅ 
5
3
​
 ⋅ 
5
2
​
 =0.192