Method I: Fix two ordered pairs of girls $AB,CD$. Then there are $7!$ ways to arrange the five boys and these two dyads. Of course we might have $ABCD$ together, so to rule that out subtract $6!$. Similarly, we must rule out another factor of $6!$ to exclude $CDAB$. Thus for these two ordered pairs there are $$7!-2\times 6!=3600$$ suitable arrangements.
Now, sticking to these pairs but varying the order gets us $$4\times 3600=14,400$$
And, finally, we can change the pairs. Instead of $AB$ we could have had $AC$ or $AD$ so, finally, $$3\times 14,400= 43,200$$
Method II: (closer to what you were trying) Arrange the kids as $$-\;AB-CD\;-$$ Where $A,B,C,D$ denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are $15$ satisfactory ways to arrange the boys: $$\{0,5,0\},\;\{1,4,0\},\;\{0,4,1\},\;\{2,3,0\},\;\{0,3,2\},\;\{1,3,1\},\;\{3,2,0\},\;\{0,2,3\},\;\{2,2,1\},\;\{1,2,2\},\;\{3,1,1\},\;\{1,1,3\},\{2,1,2\},\;\{4,1,0\},\;\{0,1,4\}$$ (I'm writing them all out because I believe you only counted $10$ of these). We must then pick some permutation of the girls to fill the slots labeled $A,B,C,D$ and pick some permutation of the boys to populate the dashed regions thus $$15\times 4!\times 5!=43,200$$