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  4. If n is an integer with 0 ≤ n ≤ 11, then the minimum value of n !(11-1) ! is attained when a value of n equals to

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Q. 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

EAMCETEAMCET 2014

Solution:

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 $n=5$.