There are 4 odd places and 3 even places.
Odd numbers are 1,1,3,3.
Odd numbers can be arranged in $\large\frac{4!}{2!.2!}$ ways.
and
even numbers are 2,2,4, and even places are 3
Even numbers can be placed in $\large\frac{3!}{2!}$ ways.
$\therefore$ The required no. of numbers = $\large\frac{4!}{2!.2!}.\frac{3!}{2!}$
$=6\times 3=18$