${ }^5 C_{x-1}>2 \cdot{ }^5 C_x$
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 $=\{5\}$