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Question Number 131994 by liberty last updated on 10/Feb/21

Commented by liberty last updated on 10/Feb/21

$$\mathrm{old}\mathrm{unswered}$$

Answered by EDWIN88 last updated on 10/Feb/21

$$\left(1\right)\mathrm{n}\left(\mathrm{S}\right)=\frac{15!}{5!.5!.5!};\mathrm{n}\left(\mathrm{A}\right)=3!\times \frac{12!}{4!.4!.4!}$$$$\mathrm{P}\left(\mathrm{A}\right)=\frac{6\times 12!\times 5!\times 5!\times 5!}{4!\times 4!\times 4!\times 15!}=\frac{25}{91}$$

Commented by liberty last updated on 10/Feb/21

$$\mathrm{Vielen}\mathrm{Dank}\mathrm{Sir}\mathrm{Edwin}$$

Answered by EDWIN88 last updated on 10/Feb/21

$$\left(2\right)\mathrm{n}\left(\mathrm{A}\right)=3\times {\mathrm{C}}_2^{12}\times {\mathrm{C}}_5^{10}\times {\mathrm{C}}_5^5=\frac{3\times 12!}{2!5!.5!}$$$$\mathrm{P}\left(\mathrm{A}\right)=\frac{3\times 12!\times 5!\times 5!\times 5!}{2\times 5!\times 5!\times 15!}=\frac{3\times 12!\times 120}{2\times 15\times 14\times 13\times 12!}$$$$=\frac{3\times 60}{15\times 14\times 13}=\frac{12}{14\times 13}=\frac{6}{91}$$

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