The solution set of ${ }^5 C_{x-1}2 \cdot\left({ }^5 C_x\right)$ is

Question

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

Aicharya · Accepted Answer

${ }^5 C_{x-1}>2 \cdot{ }^5 C_x$ Inequality defines, when $x-1>0 ; \therefore x-1$ and $x$ are positive integers $ \begin{aligned} & x>0 \\ & x-1 < 5 \Rightarrow x < 5 \end{aligned} $ Now, ${ }^5 C_{x-1}>2 \cdot{ }^5 C_x$ $ \begin{array}{cc} & \frac{{ }^5 C_{x-1}}{{ }^5 C_x}>2 \\ & \left.\frac{5 !}{\frac{(5-x+1) !(x-1) !}{(5-x) ! x !}>2}{ }^{n !} C_r=\frac{n !}{(n-r) ! r !}\right\} \\ \Rightarrow & \frac{(5-x) ! x(x-1)}{(5-x+1)(5-x) !(x-1) !}>2 \\ \Rightarrow & \frac{x}{5-x+1}>2 \\ \Rightarrow & \frac{x-12+2 x}{6-x}>0 \\ \Rightarrow & \frac{3 x-12}{6-x}>0 \Rightarrow \frac{3(x-4)}{x-6} < 0 \end{array} $ Critical points are $x=4,6$

$\therefore x \in I^{+} \Rightarrow x=5$ is only one possible value. $\therefore$ Solution set $=\{5\}$