Question

The least positive integral value of $x$ which satisfies the in equality ${}^{10}{C}_{x-1}>{2.}^{10}{C}_x$ is

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

Answer: (B)

$8$

For above expression to be defined we must have,
$10\ge x-1\Rightarrow x\le 11$
And $10\ge x\Rightarrow x\le 10$
Given inequation is
${}^{10}{C}_{x-1}>{2.}^{10}{C}_x$
$\Rightarrow 1\ge 2\cdot \frac{{}^{10}{C}_x}{{}^{10}{C}_{x-1}}$
$\Rightarrow 1>2.\frac{10-x+1}{x},\because \frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{n-r+1}{r}$
$\Rightarrow x>2(11-x)\Rightarrow 3x>22$
$\Rightarrow x>22/3$
Using above we get $x\in (22/3,10]$
But since ${}^n{C}_r$ is only defined for non-negative integral values of $n$ and $r$
$\Rightarrow x=8,9$
Hence, least positive value is $8$.