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I'm working on this problem:

An elevator in a building starts with 5 passengers and stops at seven 7 floors. If each passenger is equally likely to get an any floor and all the passengers leave independently of each other, what is the probability that no two passengers will get off at the same floor?

I figure you could do 7 P 5 - since the first guy has 7 floors he can get off on, the second 6, etc. But do I need to account for the order that the people can choose their floor, and therefore multiply that by 5 factorial? Or am I approaching this incorrectly in the first place?

Thanks

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The number of ways to assing $7$ floors to $5$ passengers is $7 \cdot 7 \cdot 7 \cdot 7 \cdot7$ beacuse for each passenger you can choose one of the $7$ floors.

The number of ways to assign $7$ floors to $5$ passengers without repetition of floors is $7 \cdot 6 \cdot5\cdot 4\cdot3$ because for the first passenger you have $7$ option, for the second you will have $6$ and so on. Note that this number count all possible orders betwen passengers too.

Then, you will guess it, the result is

$$\frac{7 \cdot 6 \cdot5\cdot 4\cdot3}{7^5}.$$

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Call the people $A$ to $E$. Consider the floors where they get off. It doesn't matter where $A$ gets off. The probability $B$ gets off at a different floor is $\dfrac{6}{7}$.

Given that $A$ and $B$ get off at different floors, the probability $C$ gets off at a different floor than $A$ and $B$ is $\dfrac{5}{7}$.

So the probability that $A,B,C$ all get off at different floors is $\dfrac{6}{7}\cdot\dfrac{5}{7}$.

Given they all got off at different floors, the probability $D$ gets off at a different floor from $A,B,C$ is $\dfrac{4}{7}$. Continue. Soon we get to $$\frac{6}{7}\cdot\frac{5}{7}\cdot\frac{4}{7}\cdot\frac{3}{7}.$$

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By way of a general formulation - if there are $k$ people and $n$ floors (where $ n \ge k$):

If we want to calculate the probability that all $k$ people go to the same floor, observe that there are only $n$ possible options, where each person can land on the same floor. So probability that all $k$ people land on the same floor is $$ \frac{n}{\underbrace{n\cdot n\cdot n\dots}_\text{$k$ times $n$}}\text{ or }\frac{n}{n^k} = \frac{1}{n^{k-1}}. $$ By law of probabilities - the probability that at least one pair of persons out of $k$ [$k \ge 2$] does NOT land on the same floor is $1 - 1/n^{k-1}$.

Also probability that no pair of persons lands on the same floor requires you to observe that the first person has $n$ choices, 2nd person has $(n-1)$ choices and so on till the $k$-th person who has $(n-k+1)$ choices and therefore the required probability is $$ \frac{n\cdot(n-1)\cdot(n-2)\dots(n-k+1)}{n^k} $$

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  • $\begingroup$ Thanks for improving the answer @taroccoesbrocco $\endgroup$ – Stats_Lover Oct 23 '18 at 2:16
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You calculate the possibilities of them not getting of at the same floor.

  • The first person can get off at any floor, there are 7 possibilities
  • The second person can get off at any floor other than the floor the first person got off at, there are 6 possibilities
  • The third person can get off at any floor other than the floor the first two people got off at, there are 5 possibilities
  • The fourth person can get off at any floor other than the floor the first three people got off at, there are 4 possibilities
  • The fifth person can get off at any floor other than the floor the first four people got off at, there are 3 possibilities

There are $7*6*5*4*3=2520$ possibilities that they dont get off on the same floor

Now you calculate the total number of possibilities for them to get off. All the people can get off at any floor, there are $7^5=16807$ possibilities

The probability that no two passengers will get off at the same floor is the first possibility divided by the second, which is $\dfrac{2520}{16807}=\dfrac{360}{2401}$ which is approximately 15%

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