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I pick $N$ numbers randomly from the set $\{1,.., N\}$ (with replacement), then what is the probability that exactly $k$ distinct numbers are chosen?

I.e. probability that $1$ distinct number is chosen would be $\left(\frac{1}{N}\right)^{N-1}$. The probability that $N$ distinct numbers are chosen would be $\frac{N!}{N^N}$

Any help would be appreciated!

user127.0.0.1
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puzzledbyprobabilities
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  • The probability that there is 1 distinct number chosen is ${N \choose 1} (1/N)^{N-1}$, since there are $N$ numbers that could be chosen. This gives a hint on how to work out other values. – Matthew Conroy Jan 13 '14 at 21:49
  • Matthew, you are saying that when $N=2$, the probability of selecting 1 distinct number is $2*\frac{1}{2} = 1$? – puzzledbyprobabilities Jan 13 '14 at 23:44
  • Oops. Yes, it should be ${N \choose 1}(1/N)^N$. Thanks. This gives a form that can be extended, using inclusion-exclusion. – Matthew Conroy Jan 13 '14 at 23:48
  • Here's an example of a way to do it directly: http://math.stackexchange.com/questions/550675/the-number-of-rolls-of-6-dice-with-exactly-4-distinct-values – Matthew Conroy Jan 13 '14 at 23:56
  • This answer also gives some ideas: http://math.stackexchange.com/questions/89794/count-the-number-of-ways-of-four-distinct-numbers-showing-up-when-six-dice-are-r – Matthew Conroy Jan 14 '14 at 01:29
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    I've discovered a complete formula in terms of Stirling numbers of the second kind here: http://math.stackexchange.com/questions/408983/expected-number-of-repeats-in-random-strings-from-different-sized-alphabets?rq=1 – puzzledbyprobabilities Jan 14 '14 at 20:17

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