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Question Number 164429 by malwan last updated on 17/Jan/22

$$IfC(2x,x)=70$$$$provethatx=4$$

Answered by alephzero last updated on 17/Jan/22

$$C_x^{2x}=70$$$$C_x^{2x}=C_4^8=\frac{8!}{4!(8-4)!}=\frac{8!}{4!4!}=$$$$=\frac{\cancel{2\times 3\times 4}\times 5\times \cancel{6}\times 7\times \cancel{8}}{\cancel{\left(2\times 3\times 4\right)}\left(2\times \cancel{3}\times \cancel{4}\right)}=$$$$=\frac{5\times 2\times 7\times \cancel{2}}{\cancel{2}}=5\times 2\times 7=10\times 7=$$$$=70$$$$\Rightarrow C_x^{2x}=70\iff x=4$$

Commented by malwan last updated on 17/Jan/22

$$directproofplease$$

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