Question

The least positive integral value of $x$ which satisfies the inequality ${}^{10}{C}_{x-1}>2\cdot {}^{10}{C}_x$ is

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is B $8$

${}^{10}{C}_{x-1}>{2.}^{10}{C}_x$
$\frac{10!}{(x-1)!(10-x+1)!}>2\times \frac{10!}{x!(10-x)!}$
$\frac{1}{(x-1)!(11-x)!}>\frac{2}{x(x-1)!(10-x)!}$
$\frac{1}{(11-x)(10-x)!}>\frac{2}{x(10-x)!}$
$\frac{1}{11-x}>\frac{2}{x}$
$x>22-2x$

$3x>22$
$x>\frac{22}{3}$
$x>7.33$
$x\ge 8$
$\therefore$ Least integral value of $x=8$