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If ${ }^{2 \mathrm{n}} C_3:{ }^{\mathrm{n}} C_3=12: 1$, then $\mathrm{n}=$
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The correct answer is:
$5$
$\frac{{ }^{2 n} C_3}{{ }^n C_3}=\frac{12}{1}$
$\frac{\frac{2 n !}{3 !(2 n-3) !}}{\frac{n !}{3 !(n-3) !}}=\frac{12}{1}$
$\begin{aligned} & \Rightarrow \frac{2 n(2 n-1)(2 n-2)}{n(n-1)(n-2)}=\frac{12}{1} \\ & \Rightarrow \frac{2 n(2 n-1) \cdot 2(n-1)}{n(n-1)(n-2)}=\frac{12}{1} \\ & \Rightarrow \frac{2 n-1}{n-2}=\frac{3}{1} \\ & \Rightarrow 2 n-1=3 n-6 \\ & \Rightarrow n=5\end{aligned}$
$\frac{\frac{2 n !}{3 !(2 n-3) !}}{\frac{n !}{3 !(n-3) !}}=\frac{12}{1}$
$\begin{aligned} & \Rightarrow \frac{2 n(2 n-1)(2 n-2)}{n(n-1)(n-2)}=\frac{12}{1} \\ & \Rightarrow \frac{2 n(2 n-1) \cdot 2(n-1)}{n(n-1)(n-2)}=\frac{12}{1} \\ & \Rightarrow \frac{2 n-1}{n-2}=\frac{3}{1} \\ & \Rightarrow 2 n-1=3 n-6 \\ & \Rightarrow n=5\end{aligned}$
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