Question

If ${}^{n-1}{C}_3{+}^{n-1}{C}_4{>}^n{C}_3$ the $n$ is just greater than

• A. $5$
• B. $6$
• C. $4$
• D. $7$

Answer 1

The correct option is D

$7$

Explanation for correct option

Given, ${}^{n-1}{C}_3{+}^{n-1}{C}_4{>}^n{C}_3$

$$\begin{gathered} \Rightarrow {}^n{C}_4>{}^n{C}_3\left[\because {}^{n}C_m+{}^{n}C_{m+1}={}^{n+1}{C}_{m+1}\right] \\ \Rightarrow \frac{n!}{4!\left(n-4\right)!}>\frac{n!}{3!\left(n-3\right)!}\left[\because {}^{n}C_m=\frac{n!}{m!\left(n-m\right)!}\right] \\ \Rightarrow \frac{1}{4}>\frac{1}{\left(n-3\right)} \\ \Rightarrow n-3>4 \\ \Rightarrow n>7 \end{gathered}$$

Hence, option (D) i.e. $7$ is correct