Question

The least positive integral value of $x$ which satisfies the inequality ${}^{10}C_{x-1}>2{}^{10}C_x$ is

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is B

$8$

Solve the given expression

Given expression,${}^{10}C_{x-1}>2{}^{10}C_x$

For above expression to be defined we must have,

From LHS,

$$\begin{gathered} \Rightarrow 10\ge x-1 \\ x\le 11 \end{gathered}$$

From RHS,

$$\begin{gathered} \Rightarrow 10\ge x \\ x\le 10 \end{gathered}$$

Now given equation,

$$\begin{gathered} {}^{10}C_{x-1}>2{}^{10}C_x \\ \Rightarrow 1>2\frac{{}^{10}C_x}{{}^{10}C_{x-1}} \\ \Rightarrow 1>2\frac{10-x+1}{x} \\ \Rightarrow x>2(11-x) \\ \Rightarrow 3x>22 \\ \Rightarrow x>\frac{22}{3} \\ \Rightarrow x>7.88 \end{gathered}$$ $[\because \frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{n}]$

Using above we get ,$x\in (\frac{22}{3},10]$

But since ${}^{n}C_r$ is only defined for non-negative integral values of $n$ and $r$

$\therefore x=8$

Therefore, the correct answer is option (B), $8$.