Question

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

• A. $n\ge 3$
• B. $n>5$
• C. $n>7$
• D. $noneofthese$

Answer 1

The correct option is C $n>7$

$n^{n-1}{C}_3+n-1C_4>{}^n{C}_3$

$n^{-1}{C}_3+n-1C_4={}^n{C}_4\therefore ({}^n{C}_{\gamma -1}+{}^n{C}_{\gamma }={}^{n+1}{C}_{\gamma })$

$n_4>{}^n{C}_3$

$\frac{n!}{(n-4)!4!}>\frac{n!}{(n-3)!3!}$

$\frac{(n-3)(n-4)!}{(n-4)!}>\frac{4!}{3!}$

$n-3>4$

$n>7$