The solution set of { }^5 C_{x-1} 2 cdot { | vedclass

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(C) Given inequality: ${ }^5 C_{x-1} > 2 \cdot { }^5 C_x$For the combinations to be defined,we must have $0 \le x-1 \le 5$ and $0 \le x \le 5$. Thus,$

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$\{5\}$

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(C) Given inequality: ${ }^5 C_{x-1} > 2 \cdot { }^5 C_x$For the combinations to be defined,we must have $0 \le x-1 \le 5$ and $0 \le x \le 5$. Thus,$x \in \{1, 2, 3, 4, 5\}$.Using the formula ${ }^n C_r = \frac{n!}{r!(n-r)!}$,we have:$\frac{{ }^5 C_{x-1}}{{ }^5 C_x} > 2$$\frac{5!}{(x-1)!(5-(x-1))!} \cdot \frac{x!(5-x)!}{5!} > 2$$\frac{x!(5-x)!}{(x-1)!(6-x)!} > 2$$\frac{x(x-1)!(5-x)!}{(x-1)!(6-x)(5-x)!} > 2$$\frac{x}{6-x} > 2$$\frac{x}{6-x} - 2 > 0$$\frac{x - 2(6-x)}{6-x} > 0$$\frac{x - 12 + 2x}{6-x} > 0$$\frac{3x - 12}{6-x} > 0$$\frac{3(x-4)}{-(x-6)} > 0$$\frac{x-4}{x-6} This inequality holds for $4 Since $x$ must be an integer,the only possible value is $x = 5$.Therefore,the solution set is $\{5\}$.

The solution set of ${ }^5 C_{x-1} > 2 \cdot { }^5 C_x$ is

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