I know the basic set up of this problem. Because each child gets at least two but no more than 4 pens for each child there is a factor equal to:
$$(x^2+x^3+x^4)^3$$ Note that I included it raised to the third power because there are three children. Now I have to find $x^8$. The question is how do I do that given this information?
Notice that André Nicolas's suggestion of giving what is surely needed first is exactly the same as Alexey Burdin's suggestion of factoring out $(x^2)^3$. And André's method should also allow you to see that the problem reduces to a much simpler one which has a standard easy solution by placing 2 dividers between the pens, since it is from the start impossible for any child to get more than 4. In general, one would use the inclusion-exclusion principle if the upper bounds are still there after getting rid of the lower bounds.