# Find the probability that each person exits an elevator on a different floor.

I have a question and I am working on to study for a test. I have a hard time deciding when to use certain techniques.

Suppose you have $5$ people on an elevator that stops at $5$ floors. Each person has an equal probability of going to any one floor. Find the probability that they all get off on different floors.

I would want to solve this by saying there are five possible floors to get off on and we want one person to get off on each. Thus: $$\frac{1^5}{5^5}$$

however the answer is $\frac{5!}{5^5}$. I do not see exactly why they got this but I assume it was solved as:

$$\frac{5}{5}\frac{4}{5}\frac{3}{5}\frac{2}{5}\frac{1}{5} = \frac{5!}{5^5}$$

Why is it done like this versus the first way which is the way I thought about solving this.

• Not following your calculation. Person $A$ can get off anywhere. Person $B$ has $4$ choices, and so on. – lulu Sep 30 '17 at 21:03
• What you calculated is the probability that all five people get off on a particular floor. – N. F. Taussig Sep 30 '17 at 21:04
• @N.F.Taussig or perhaps that Person A gets off on the first floor, Person B gets off on the second floor, ... – Henry Oct 1 '17 at 8:39
• @Henry Well spotted. In any case, the point is that Derek did not account for the number of ways five people could exit the elevator on five different floors. – N. F. Taussig Oct 1 '17 at 8:41

Let's get the denominator first. There are exactly five ways (five floors) each passenger can get off the elevator. Passenger #1 can get off at any one of five floors. Same for the other $4$ passengers. So the denominator is $5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 5^5$.
The numerator is the number of ways (count) one and only one passenger exits at each floor. So this is simply $5C1 \cdot 4C1 \cdot 3C1 \cdot 2C1 \cdot 1C1$ or $5!$, where $nCr$ is "$n$ choose $r$" or in our case $5C1$ = "$5$ choose $1$."
So answer is $5!/5^5$.