The least positive integral value of x for which 10Cx−1>2 10Cx

Solution:

$\begin{array}{l} ^{10} C_{x - 1} > 2 (^{10} C_{x}) \\ \Rightarrow \frac{10!}{(x - 1)! (11 - x)!} > 2 \frac{10!}{x! (10 - x)!} \\ \Rightarrow x > 2 (11 - x) \Rightarrow 3 x > 22 \\ \Rightarrow x > 7 \Rightarrow x \geq 8 . \end{array}$

Thus, least positive integral value of $x$ is $8$

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