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If $n$ is an integer with $0 \leq n \leq 11$, then the minimum value of $n !(11-1)$ ! is attained when a value of $n$ equals to
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The correct answer is:
$5$
We know, ${ }^{11} C_n$ is maximum when $n=5$.
$$
\therefore{ }^{11} C_n=\frac{11 !}{n !(11-n) !}
$$
$\therefore n !(11-n) !$ is minimum when $n=5$.
$$
\therefore{ }^{11} C_n=\frac{11 !}{n !(11-n) !}
$$
$\therefore n !(11-n) !$ is minimum when $n=5$.
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