The solution set of ^{10}C_{x-1} 2 cdot ^{1 | vedclass

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(C) Given inequality: $^{10}C_{x-1} > 2 \cdot ^{10}C_x$ Using the formula $^{n}C_r = \frac{n!}{r!(n-r)!}$,we have: $ \frac{10!}{(x-1)!(10-(x-1))!} > 2

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$ \{8, 9, 10\} $

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(C) Given inequality: $^{10}C_{x-1} > 2 \cdot ^{10}C_x$ Using the formula $^{n}C_r = \frac{n!}{r!(n-r)!}$,we have: $ \frac{10!}{(x-1)!(10-(x-1))!} > 2 \cdot \frac{10!}{x!(10-x)!} $ $ \frac{1}{(x-1)!(11-x)!} > \frac{2}{x!(10-x)!} $ Since $x! = x \cdot (x-1)!$ and $(11-x)! = (11-x) \cdot (10-x)!$,we get: $ \frac{1}{(11-x)(10-x)!} > \frac{2}{x(10-x)!} $ $ \frac{1}{11-x} > \frac{2}{x} $ Since $x$ must be a positive integer such that $1 \le x \le 10$,we have $x > 22 - 2x \implies 3x > 22 \implies x > 7.33$. Thus,the possible integer values for $x$ are $8, 9, 10$. Therefore,the solution set is $\{8, 9, 10\}$.

The solution set of $^{10}C_{x-1} > 2 \cdot ^{10}C_x$ is

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