Distributing identical objects to different people. What is the number of ways in which we can distribute 12 identical oranges among 4 children such that every child gets at least one and no child gets more than 4 ?
Till now ,My attempts have focused on first finding out all such ways in each person gets at least one orange by ${ 13 \choose 3}$ and then trying to find out number of those ways in which at least one person gets more than 4 oranges. 
( Basically trying to apply to the inclusion- exclusion principle ) 
However, I have not been able to figure out the former ( i.e. number of ways in which each person gets more than 4 oranges ). 
Is this approach the most suitable one , or should I look at something different ?

The number of ways $12$ oranges can be distributed to four children if each child gets at least one orange is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 12 \tag{1}$$
in the positive integers.  A particular solution of equation 1 in the positive integers corresponds to the placement of three addition signs in the eleven spaces between successive ones in a row of $12$ ones. For instance,
$$1 + 1 1 + 1 1 1 + 1 1 1 1 1 1$$
corresponds to the solution $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, and $x_4 = 6$.  Hence, the number of solutions of equation 1 in the positive integers is 
$$\binom{11}{3}$$
Since each child receives at most four oranges, we must exclude those solutions in which one or more of the variables exceeds $4$.  Since $3 \cdot 5 = 15 > 12$, at most two of the variables can exceed $4$ simultaneously.
Suppose $x_1 > 4$.  Let $y_1 = x_1 - 4$.  Then $y_1$ is a positive integer.  Substituting $y_1 + 4$ for $x_1$ in equation 1 yields
\begin{align*}
y_1 + 4 + x_2 + x_3 + x_4 & = 12\\
y_1 + x_2 + x_3 + x_4 & = 8 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers with $\binom{7}{3}$ solutions.  By symmetry, there are $\binom{7}{3}$ solutions in which one of the four variables exceeds $4$.  Hence, there are 
$$\binom{4}{1}\binom{7}{3}$$
solutions in which one variable exceeds $4$.
However, subtracting $\binom{4}{1}\binom{7}{3}$ from $\binom{11}{3}$ removes those solutions in which two of the variables exceed $4$ twice.  Since we only want to remove such solutions once, we must add the number of solutions in which two of the variables exceed $4$.
Suppose $x_1$ and $x_2$ exceed $4$.  Let $y_1 = x_1 - 4$; let $y_2 = x_2 - 4$.  Then $y_1$ and $y_2$ are positive integers.  Substituting $y_1 + 4$ for $x_1$ and $y_2 + 4$ for $x_2$ in equation 1 yields 
\begin{align*}
y_1 + 4 + y_2 + 4 + x_3 + x_4 & = 12\\
y_1 + y_2 + x_3 + x_4 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the positive integers with one solution (each variable is equal to $1$).  By symmetry, there is one solution for each of the $\binom{4}{2}$ ways in which two of the variables exceed $4$, so there are
$$\binom{4}{2}\binom{3}{3}$$
solutions in which two of the variables exceed $4$.  
By the Inclusion-Exclusion Principle, the number of ways the twelve oranges can be distributed to four children so that each child receives at least one and at most four oranges is 
$$\binom{11}{3} - \binom{4}{1}\binom{7}{3} + \binom{4}{2}\binom{3}{3}$$
