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Suppose that I have $n$ objects and I make $m$ choices (with repetitions) out of the objects. Then what is the probability that no two of my choices are the same?

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The total number of ways you can make $m$ choices with repetitions is $$\underbrace{n \times n \times \cdots \times n}_{m \text{ times}}$$ Now if want no two of the $m$ choices to be the same, then the number of ways is $$n \times (n-1) \times (n-2) \times \cdots \times (n-m+1)$$ Hence, the desired probability is $$\dfrac{n \times (n-1) \times (n-2) \times \cdots \times (n-m+1)}{n^m} = \dfrac{n!}{(n-m)!} \dfrac1{ n^m}$$

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