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If 12 balls are thrown at random into 20 boxes, what is the probability that no box will receive more than 1 ball?

I can understand the answer: $$\frac {{}_{20}P_{12}}{20^{12}}$$

Then I try using stars and bars to get the dominator: $$\frac {{}_{20}C_{12}}{{}_{12+20-1}C_{20-1}}$$

What's wrong with the second way?

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  • $\begingroup$ Welcome to MSE. Please use MathJax to format your question. $\endgroup$
    – saulspatz
    Commented Feb 9, 2019 at 15:29
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    $\begingroup$ You can't use stars and bars because that counts the number of different distributions, but does not take account of their relative likelihood. The distributions where all balls land in the first box is counted the same as the one where the first $8$ boxes have $2$ balls apiece and the last $4$ have $1$ each, but the second distribution is much more probable. $\endgroup$
    – saulspatz
    Commented Feb 9, 2019 at 15:32

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