# Probability of combinations

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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Any restriction on M and N ? –  Quixotic Apr 21 '11 at 20:24

$\frac{N(N-1)\dots(N-M+1)}{N^M}$
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. –  svenkatr Apr 21 '11 at 20:30
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 $M>N$. –  Henry Apr 21 '11 at 20:53