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The least value of $n$ for which ${ }^{(n-1)} C_2+{ }^{(n-1)} C_3>{ }^n C_2$ is
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The correct answer is:
$6$
${ }^{n-1} C_2+{ }^{n-1} C_3>{ }^n C_2$
${ }^n C_3>{ }^n C_2 \Rightarrow \frac{n !}{3 !(n-3) !}>\frac{n !}{2 !(n-2) !}$
$\Rightarrow \frac{1}{3} \geq \frac{1}{(n-2)}$
$\begin{aligned} & \Rightarrow \mathrm{n}-2>3 \Rightarrow \mathrm{n}>5 \\ & \Rightarrow \mathrm{n}=6\end{aligned}$
${ }^n C_3>{ }^n C_2 \Rightarrow \frac{n !}{3 !(n-3) !}>\frac{n !}{2 !(n-2) !}$
$\Rightarrow \frac{1}{3} \geq \frac{1}{(n-2)}$
$\begin{aligned} & \Rightarrow \mathrm{n}-2>3 \Rightarrow \mathrm{n}>5 \\ & \Rightarrow \mathrm{n}=6\end{aligned}$
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