# Quod Erat Demonstrandum

## 2011/08/17

### 還是數算

Filed under: Additional / Applied Mathematics,HKALE,NSS,Teaching — johnmayhk @ 11:29 下午

There are 10 empty boxes. 5 balls are going to put one by one into a randomly selected box. Find the probability that two of the boxes each contains 2 balls.

$\frac{9}{10}\times \frac{8}{10}\times \frac{3}{10}\times \frac{2}{10}$

$\frac{9}{10}\times \frac{8}{10}\times \frac{3}{10}\times \frac{2}{10}$ = $\frac{10\times 9\times 8\times 3 \times 2}{10^5}$

$\frac{C_2^{10}\times C_2^5\times C_2^3\times 8}{10^5}$

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$C_2^{10}\times 8\times C_2^5\times C_2^3$

$\frac{C_2^{10}\times C_2^5\times C_2^3\times 8}{10^5} = \frac{9}{10}\times \frac{8}{10}\times \frac{3}{10}\times \frac{2}{10} \times \frac{5}{2}$

$\frac{10}{10}\times \frac{1}{10}\times \frac{9}{10}\times \frac{1}{10}\times \frac{8}{10}$