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The least value of a natural number $n$ such that $\left(\begin{array}{c}n-1 \\ 5\end{array}\right)+\left(\begin{array}{c}n-1 \\ 6\end{array}\right) < \left(\begin{array}{l}n \\ 7\end{array}\right)$, where $\left(\begin{array}{l}n \\ r\end{array}\right)=\frac{n !}{(n-r) ! r !}$, is
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14
${ }^{n-1} C_{5}+{ }^{n-1} C_{6} < { }^{n} C_{7}$
${ }^{n} C_{6} < { }^{n} C_{7}$
$\frac{n !}{6 !(n-6) !} < \frac{n !}{7 !(n-7) !}$
$n-6>7$
$n>13$
$n_{\min }=14$
${ }^{n} C_{6} < { }^{n} C_{7}$
$\frac{n !}{6 !(n-6) !} < \frac{n !}{7 !(n-7) !}$
$n-6>7$
$n>13$
$n_{\min }=14$
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