1234

$2,4,2,\quad can\quad be\quad arrenged\quad in\quad three\quad even\quad places\quad in\frac { 3! }{ 2! } =3\quad ways\\ \quad and\quad 1,3,3,1\quad can\quad be\quad arrenged\quad in\quad four\quad odd\quad places\quad in\quad \frac { 4! }{ 2!2! } =6\quad ways.\\ \quad So\quad required\quad number\quad of\quad numbers\quad is\quad 3\times 6=18$

Here are the $6$ possible odd places of the seven digit number

1 _ 3 _ 3 _ 1

1 _ 3 _ 1 _ 3

1 _ 1 _ 3 _ 3

3 _ 1 _ 1 _ 3

3 _ 1 _ 3 _ 1

3_ 3 _ 1 _ 1

Here are the $3$ possible combinations of even places for every combination above of odd places

_ 2 _ 4 _ 2 _

_ 2 _ 2 _ 4 _

_ 4 _ 2 _ 2 _

Giving us $3*6 = \boxed{18}$ total possibilities.