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

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(C) The given inequality is $^{10}C_{x-1}  >  2 \cdot ^{10}C_x$.Using the formula $^{n}C_r = \frac{n!}{r!(n-r)!}$, we expand the terms:$\frac{10!}{(x-

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{$8$,$9$,$10$}

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(C) The given inequality is $^{10}C_{x-1}  >  2 \cdot ^{10}C_x$.Using the formula $^{n}C_r = \frac{n!}{r!(n-r)!}$, we expand the terms:$\frac{10!}{(x-1)!(10-(x-1))!}  >  2 \cdot \frac{10!}{x!(10-x)!}$$\frac{10!}{(x-1)!(11-x)!}  >  2 \cdot \frac{10!}{x!(10-x)!}$Since $10!$ is common, we simplify:$\frac{1}{(x-1)!(11-x)(10-x)!}  >  2 \cdot \frac{1}{x(x-1)!(10-x)!}$$\frac{1}{11-x}  >  \frac{2}{x}$Since $x$ must be a positive integer and $1 \le x \le 10$, we cross-multiply:$x  >  2(11-x)$$x  >  22 - 2x$$3x  >  22$$x  >  \frac{22}{3} \approx 7.33$Since $x$ is an integer and $x \le 10$, the possible values for $x$ are ${8, 9, 10}$.

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

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