Dance couples riddle

Imagine there are $5$ women and $5$ men on a disco. Two of the gals have two brothers within the guys.

(image shows sibling relations)

In how many different ways can female-male couples be arranged if siblings are not allowed to dance together?

Now, I've cheated a bit and got the answer, $48$ different combinations, using a computer program. But I'm having trouble trying to find a more elegant and general mathematical solution to this problem.

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If she decides to dance with one of the two brothers of the other girl, the other girl has $3$ options. The number of ways the other three girls can pair up with the other three guys is $3!$. Hence, the number of ways is $2 \times 3 \times 3!$.
If the first girl decides to dance with the guy who doesn't have a sister, then the second girl has only $2$ options. And as above, the number of ways the other three girls can pair up with the other three guys is $3!$. Hence, the number of ways in this case is $1 \times 2 \times 3!$.
Adding both gives us $(2 \times 3 + 2) \times 3! = 8 \times 6 = 48$.