So the way I am trying to solve this is as follows:
- Allow the girls to be arranged first : $3!$
- There are 4 spaces that are created for the three boys : $4P3$ (since the order is important $4C3 \times 3!$, which is $4P3$)
Therefore total possibilities $=3! \times 4P3 = 144$
Now the answer to the problem is $72$, which is half of my answer. What am I double counting?