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${ }^{2 \mathrm{n}} \mathrm{C}_4:{ }^{\mathrm{n}} \mathrm{C}_3=99: 4 \Rightarrow \mathrm{n}=$
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The correct answer is:
$6$
${ }^{2 n} C_4:{ }^n C_3=99: 4$
$\Rightarrow \frac{\frac{2 n !}{4 !(2 n-4) !}}{\frac{n !}{3 !(n-3) !}}=\frac{99}{4}$
$\Rightarrow \frac{2 \mathrm{n}(2 \mathrm{n}-1)(2 \mathrm{n}-2)(2 \mathrm{n}-3)}{4 \cdot \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}=\frac{99}{4}$
$\Rightarrow \frac{2(2 n-1) \cdot 2(2 n-3)}{(n-2)}=99$
$\begin{aligned} & \Rightarrow 4\left(4 n^2-6 n-2 n+3\right)=99 n-198 \\ & \Rightarrow 16 n^2-32 n+12=99 n-198 \\ & \Rightarrow 16 n^2-131 n+210=0 \Rightarrow(n-6)(16 n-35)=0 \\ & \Rightarrow n-6=0 \text { or } 16 n-35=0\end{aligned}$
$\Rightarrow \mathrm{n}=6$ or $\mathrm{n}=\frac{35}{16}$ (Not possible)
$\Rightarrow \frac{\frac{2 n !}{4 !(2 n-4) !}}{\frac{n !}{3 !(n-3) !}}=\frac{99}{4}$
$\Rightarrow \frac{2 \mathrm{n}(2 \mathrm{n}-1)(2 \mathrm{n}-2)(2 \mathrm{n}-3)}{4 \cdot \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}=\frac{99}{4}$
$\Rightarrow \frac{2(2 n-1) \cdot 2(2 n-3)}{(n-2)}=99$
$\begin{aligned} & \Rightarrow 4\left(4 n^2-6 n-2 n+3\right)=99 n-198 \\ & \Rightarrow 16 n^2-32 n+12=99 n-198 \\ & \Rightarrow 16 n^2-131 n+210=0 \Rightarrow(n-6)(16 n-35)=0 \\ & \Rightarrow n-6=0 \text { or } 16 n-35=0\end{aligned}$
$\Rightarrow \mathrm{n}=6$ or $\mathrm{n}=\frac{35}{16}$ (Not possible)
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