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.

The question reads:

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.

  • 1
    $\begingroup$ Not following your calculation. Person $A$ can get off anywhere. Person $B$ has $4$ choices, and so on. $\endgroup$ – lulu Sep 30 '17 at 21:03
  • $\begingroup$ What you calculated is the probability that all five people get off on a particular floor. $\endgroup$ – N. F. Taussig Sep 30 '17 at 21:04
  • 1
    $\begingroup$ @N.F.Taussig or perhaps that Person A gets off on the first floor, Person B gets off on the second floor, ... $\endgroup$ – Henry Oct 1 '17 at 8:39
  • $\begingroup$ @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. $\endgroup$ – 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 $5!/5^5$.

| cite | improve this answer | |
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 1 '17 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.