If { }^{2n}C_3 : { }^{n}C_3 = 12 : 1,then n = | vedclass

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(A) Given $\frac{{ }^{2n}C_3}{{ }^{n}C_3} = \frac{12}{1}$.Using the formula ${ }^{n}C_r = \frac{n!}{r!(n-r)!}$,we have:$\frac{\frac{(2n)!}{3!(2n-3)!}}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given $\frac{{ }^{2n}C_3}{{ }^{n}C_3} = \frac{12}{1}$.Using the formula ${ }^{n}C_r = \frac{n!}{r!(n-r)!}$,we have:$\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 12$$\Rightarrow \frac{2n(2n-1)(2n-2)(2n-3)!}{3!(2n-3)!} 	imes \frac{3!(n-3)!}{n(n-1)(n-2)(n-3)!} = 12$$\Rightarrow \frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 12$$\Rightarrow \frac{4n(2n-1)}{n(n-2)} = 12$$\Rightarrow \frac{4(2n-1)}{n-2} = 12$$\Rightarrow 2n-1 = 3(n-2)$$\Rightarrow 2n-1 = 3n-6$$\Rightarrow n = 5$.

If ${ }^{2n}C_3 : { }^{n}C_3 = 12 : 1$,then $n =$

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