Find the coefficient of $x^{13}$ in the convolution of two generating functions 
Four thiefs have stolen a collection of 13 identical diamonds. After the
  theft, they decided how to distribute them.
3 of them have special requests: 
  
  
*
  
*One of them doesn't want more than 2 diamonds ($\leq2$).
  
*The other one only wants a number of diamonds that's a multiple of 3.
  
*And the other one wants an odd number of diamonds greater or equal than 3.
Find in how many ways they can distribute the diamonds.

My first thought was to use generating functions to find the coefficient of $x^{13}$, for this problem it would be:
$f(x)=(1+x+x^2+x^3+...)(1+x+x^2)(1+x^3+x^6+...)(x^3+x^5+x^7+...)=\frac{1}{1-x} \frac{1-x^3}{1-x} \frac{1}{1-x^3}\frac{x^3}{1-x^3}=\frac{1}{(1-x)^{2}}\frac{x^3}{(1-x^2)}$ 
and that would be equivalent to finding the coefficient of $x^{10}$ in $\frac{1}{(1-x)^{2}}\frac{1}{(1-x^2)}$.
I know that I could use the binomial theorem, but the solution I have says I should be using convolution of these two generating functions, but I have no idea how to use it to find the coefficient of $x^{10}$.

Hint:
$$\begin{align}
\frac{1}{(1-x)^2} \frac{1}{1-x^2} &= (1-x)^{-2} (1-x^2)^{-1} \\
&= \sum_{i=0}^{\infty}\binom{2+i-1}{i} x^i\cdot \sum_{j=0}^{\infty}x^{2j}
\end{align}$$
Combinations of $(i,j)$ satisfying $i+2j=10$ are $(10,0),(8,1),(6,2),(4,3),(2,4),(0,5)$. 
[The above list of combinations was corrected in an edit on  3/10/2019.  It just goes to show that I should not attempt to write an answer just before I rush out the door.  My apologies.]
Can you take it from there?
