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An elevator in a building starts with five passengers and stops at seven floors. If each passenger is equally likely to get off on 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?

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Hint: There are $7^5$ ways for the passengers to leave the elevator, but only $7\cdot6\cdot5\cdot4\cdot3$ avoid two passenges getting off at the same floor.

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  • $\begingroup$ Good enough. Thanks! $\endgroup$ – James Sep 14 '12 at 7:13
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We can rephrase the question as how many ways can a single passenger choose a floor to get off the elevator? There are five passengers and seven floors. First passenger has seven floors to choose from, second passenger has six floors to choose from, etc.

Now of course the probability is the answer to the above question over total number of ways five passengers can choose seven floors.

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    $\begingroup$ hint: Hagen's hint is more than a hint $\endgroup$ – MercSuey Sep 14 '12 at 6:19

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