Can we solve this using stars and bars? The number of ways of distributing 12 identical oranges among 4 children so that every child gets at least one and no child more than 4 is 31.
My try: First of all give each child 1 orange and we are left with 8 oranges.
Then 
3 3 2 0
3 3 1 1
3 2 2 1
2 2 2 2
and permuting each, we get total 31. But can we solve this using stars and bars?

As SandeepThilakan said in the comments above, the coefficient of $x^{12}$ in the expansion of the polynomial $(x^1+x^2+x^3+x^4)^4$ counts the way that between 1 to 4 objects distributed to 4 groups gives a total of 12 objects.
Using the binomial expansion theorem:
$$(x+x^2+x^3+x^4)^4 \\ = ((x+x^2)+x^2(x+x^2))^4 \\ = x^4(1+x)^4(1+x^2)^4 \\ = x^4 \left(\sum\limits_{i=0}^4 {4\choose i}x^i \right)\left(\sum\limits_{j=0}^4 {4\choose j}x^{2j} \right) \\= x^4 \sum\limits_{i=0}^4 \sum\limits_{j=0}^4 {4 \choose i}{4 \choose j} x^{i+2j}$$
So the coefficients of $x^{12}$ must be $${4 \choose 0}{4 \choose 4}+{4 \choose 2}{4 \choose 3}+{4 \choose 4}{4 \choose 2} = 31$$
