$\frac{10!}{(10-x+1)!*(x-1)!}<2\frac{10!}{(10-x)!*(x)!}$
$\frac{10!}{(10-x+1)*(10-x)!*(x-1)!}<2\frac{10!}{(10-x)!*(x)*(x-1))!}$
cancelling out common terms we get
$\frac{1}{(10-x+1)}$ < $\frac{2}{x}$
Equating it we get
x<$\frac{22}{3}$
so x>1 and x< 7
x={1,2,3,4,5,6,7}