Question

If ${}^{n-1}{C}_3{+}^{n-1}{C}_4{>}^n{C}_3$, then set of complete values of n is:

Answer 1

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

we,${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$

${}^{n+1}{C}_4{+}^{n+1}{C}_{4-1}{>}^n{C}_3$

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

${}^n{C}_4{>}^n{C}_3$

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

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

$\frac{1}{4}>\frac{1}{n-3}$

$\frac{1}{n-1}-\frac{1}{4}<0$

$\frac{4-n+1}{4(n-1)}<0$

$\frac{n-3}{(n-1)}>0\frac{+-+}{13}$

$xϵ(-\infty ,1)\cup (3,\infty )$