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I am encountering a situation that essentially boils down to this: There are M people. They are randomly assigned to N chairs. What is the probability that no two persons are assigned the same chair (probability that all are assigned separate chairs).

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  • $\begingroup$ Any restriction on M and N ? $\endgroup$
    – Quixotic
    Apr 21, 2011 at 20:24

2 Answers 2

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$\frac{N(N-1)\dots(N-M+1)}{N^M}$

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    $\begingroup$ Your answer should just be $\frac{N(N-1)\dots(N-M+1)}{N^M}$ because he is asking for probability that no two or more people are assigned the same chair. $\endgroup$
    – svenkatr
    Apr 21, 2011 at 20:30
  • $\begingroup$ @svenkatr: Yes, indeed. $\endgroup$
    – Emre
    Apr 21, 2011 at 20:32
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    $\begingroup$ It is another example of the Birthday problem: for $M=23$ and $N=365$ it give $0.4927\ldots$, slightly less than a half. It might be worth noting that this expression correctly gives 0 whenever $M>N$. $\endgroup$
    – Henry
    Apr 21, 2011 at 20:53
  • $\begingroup$ M=N is okay too $\endgroup$
    – Emre
    Apr 21, 2011 at 21:00
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This seems correct. For , Total number of permutations is $N^M$ where repetitions are allowed and $N(N-1)(N-2)\cdots(N-M+1)$ are permutations where repetitions are not allowed.

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