$24 ways$.
since A have to be of form either $13x $ or $14y$
if $A=14y$ see $y$ have to be atleast 2, but then $D$ have to be atmost 1. contradiction.
so, $A=13x$
so, $x+D=13$
now, consider $B=uv $ and $C=pq$ then either $u+p=13$ or $u+p=14$
if $u+p=14$ then $v+q\geq 2+4$ contradiction.
so, $u+p=13$ so, $v+q=13$
but, leaving $1,3$ we have three disjoint pairs giving total $13$ in exactly 1 way.
$4+9=5+8=6+7=13$
so, considering $B<C$ i.e. $ u<p$ we have only
$ 3\times 2\times 2!2!$ many soluions.
so, $24 $ solutions.