For each of the 3 girls to be seated between 2 boys, the minimum number of boys needed is 4. The arrangement in which this happens is:
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Since there are 5 boys, there is one boy left over, who can be placed next to any of the 4 boys already placed. So there are 4 possible arrangements if each child is identified only by their gender. They are:
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If each child is distinguishable (not identified only by their gender) then the number of possible arrangements is
(4) x (5!) x (3!)
= 4 x 120 x 6
= 2,880.