Question

If ${}^{n-1}{C}_6+{}^{n-1}{C}_7>{}^n{C}_6$, then

• A. $n>4$
• B. $n>12$
• C. $n\ge 13$
• D. $n>13$

Answer 1

Answer: (D)

${}^{n-1}{C}_6+{}^{n-1}{C}_7={}^n{C}_7={\therefore }^n{C}_7>{}^n{C}_6$.
$\Rightarrow \frac{n!}{7!n-7!}>\frac{n!}{6!n-6!}$
$\Rightarrow \frac{1}{7.6!n-7!}>\frac{1}{6!\left(n-6\right)\left(n-7\right)!}$
$\Rightarrow n-6>7$
$\Rightarrow n>13$