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).
Probability of combinations
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probability
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0Any restriction on M and N ? – 2011-04-21
2 Answers
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$\frac{N(N-1)\dots(N-M+1)}{N^M}$
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0M=N is okay too – 2011-04-21
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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.