Question

In ${}^{n-1}{C}_3+{}^{n-1}{C}_4>{}^n{C}_3$, then $n$ is just greater than integer.

• A. $5$
• B. $6$
• C. $4$
• D. $7$

Answer 1

Answer: (D)

${}^{n-1}{C}_3+{}^{n-1}{C}_4>{}^n{C}_4>{}^n{C}_3$
$\Rightarrow \frac{n!}{4!\left(n-4\right)!}>\frac{3!}{3!\left(n-3\right)!}$
$\Rightarrow \frac{1}{4}>\frac{1}{n-3}$
$\Rightarrow n-3>4$
$\Rightarrow n>7$