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- #1

#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

*Please give only a hint, and not the full solution.*

- Thread starter Alexmahone
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- Thread starter
- #1

- Jan 26, 2012

- 268

- Feb 13, 2012

- 1,704

http://mathworld.wolfram.com/MultinomialDistribution.html

Kind regards

$\chi$ $\sigma$

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- #3

- Feb 14, 2012

- 3,682

I believe we share the same attitude: Seeing the full solution is like, killing our imagination.

Please give only a hint, and not the full solution.

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- #4

- Jan 26, 2012

- 268

The 1st ball can be put into any of the 20 boxes, the 2nd ball can can be put into any of the other 19 boxes and so on. So, the number of ways to put 12 balls into 20 boxes so that no box receives more than one ball is [TEX]20*19*\cdots*9=\frac{20!}{8!}[/TEX].

The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, the total number of ways to put 12 balls into 20 boxes is [TEX]20^{12}[/TEX].

So, [TEX]P=\frac{20!}{8!20^{12}}[/TEX]