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Question Number 131876 by I want to learn more last updated on 09/Feb/21

Answered by EDWIN88 last updated on 09/Feb/21

$$\left(\mathrm{i}\right){\mathrm{C}}_8^{10}=\frac{10\times 9}{2\times 1}=45$$

Answered by liberty last updated on 09/Feb/21

$$\left(\mathrm{ii}\right){\mathrm{C}}_{8-3}^{10-3}={\mathrm{C}}_5^7=\frac{7\times 6}{2\times 1}=21$$

Answered by physicstutes last updated on 10/Feb/21

$$\mathrm{number}\mathrm{of}\mathrm{questions}\mathrm{required}\mathrm{to}\mathrm{answer}=8$$$$\mathrm{total}=10$$$$\left(\mathrm{i}\right){}^{10}{\mathrm{C}}_8=\frac{10!}{2!8!}$$$$\left(\mathrm{ii}\right){}^7{\mathrm{C}}_5=\frac{7!}{2!5!}$$$$\left(\mathrm{iii}\right)\mathrm{if}\mathrm{he}\mathrm{answers}\mathrm{first}\mathrm{question}\mathrm{or}2\mathrm{nd}\mathrm{question}={}^9{\mathrm{C}}_7$$$$\mathrm{if}\mathrm{he}\mathrm{answers}\mathrm{first}\mathrm{two}={}^8{\mathrm{C}}_6$$$$\Rightarrow \mathrm{total}=2\left({}^9{\mathrm{C}}_7\right)+{}^8{\mathrm{C}}_6$$

Commented by I want to learn more last updated on 23/Feb/21

$$\mathrm{Thanks}\mathrm{sirs}.\mathrm{I}\mathrm{appreciate}.$$

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