Question

Solve the following inequalities.
$C_6^n<C_4^n,nϵN$

Answer 1

Ques solve: $C_6^n<{}^n{C}_{4}nϵN$

Soln $C_6^n<C_4^n$

by formula $C_r^n=\frac{n!}{\left(r\right)!(n-r)!}$

$\Rightarrow \frac{n!}{(n-6)!\left(6\right)!}<\frac{n!}{(n-4)!4!}$

$\Rightarrow \frac{1}{(n-6)!\left(6\right)!}<\frac{1}{(n-4)(n-5)(n-6)!4!}$

$\Rightarrow \frac{1}{6!}\times 4!<\frac{1}{(n-4)(n-5)}$

$\Rightarrow \frac{1\times 4!}{6\times 5\times 4!}<\frac{1}{(n-4)(n-5)}$

$\Rightarrow (n-4)(n-5)<30$

$\Rightarrow n^2-9n+20-30<0$

$\Rightarrow n^2-9n-10<0$

Let $n^2-9n-10=0$ thus $n=\frac{+9\pm \sqrt{81-4(-10)}}{2}$

$=\frac{+9\pm \sqrt{81+40}}{2}$

$=\frac{9\pm \sqrt{121}}{2}=\frac{9\pm 11}{2}$

then $n=\frac{9+11}{2}$ and $n=\frac{9-11}{2}$

$=\frac{20}{2}=10$ and $n=\frac{(-2)}{2}=-1$

Now solve $(n^2-9n-10)<0$

$\Rightarrow (n-10)(n+1)<0$

$\Rightarrow nϵ(-1,10)$