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If $ ^{n-1}C_6+^{n-1}C_7>^nC_6$, then $n$ is ?

$\begin{array}{1 1} n > 11 \\ n > 12 \\ n > 13 \\ n > 14 \end{array}$

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A)
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  • $^nC_r+^nC_{r+1}=^{n+1}C_{r+1}$
Given: $ ^{n-1}C_6+^{n-1}C_7>^nC_6$
$\Rightarrow\:^nC_7>^nC_6$
$\large\frac{n!}{(n-7)!.7!}$$>\large\frac{n!}{(n-6)!.6!}$
$\Rightarrow\:n-6>7\:\Rightarrow\:n>13$
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