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If ${ }^{(n-1)} C_3+{ }^{(n-1)} C_4>{ }^n C_3$, then the minimum value of $n$ is
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Verified Answer
The correct answer is:
8
Given, ${ }^{n-1} C_3+{ }^{n-1} C_4>{ }^n C_3$
$\begin{array}{lc}
\therefore & { }^n C_4>{ }^n C_3 \\
& {\left[\because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right]} \\
\Rightarrow & n-4>3 \\
\Rightarrow & n>7 \\
\therefore & n=8
\end{array}$
$\begin{array}{lc}
\therefore & { }^n C_4>{ }^n C_3 \\
& {\left[\because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right]} \\
\Rightarrow & n-4>3 \\
\Rightarrow & n>7 \\
\therefore & n=8
\end{array}$
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