Constrained stars and bars. I have been racking my brain to understand the solutions of constrained stars and bars problems like:
$a+b+c+d = 19$;  $a,b,c,d \in \{0,1,2,3,4,5,6,7\}$
Finding the total number of solutions is really simple $\dbinom{22}{3}$. 
What to do after that? I want to understand the entire procedure step by step. 
Please do not mark it as a duplicate of this.
Reasons: 
The accepted answer's trick doesn't always work. 
Another person (the answer with two upvotes) has just listed out the binomial coefficients and I can't even remotely understand what he has tried to do. 
The second answer isn't my cup of tea. 

Here's another method which applies generally to many such problems. You have the problem equivalent to searching for coefficient of $x^{19}$ in expansion of $(1+x+x^2 ... x^7)^4$
Each power of $x$ in each of the $4$ factors represents value of particular variable ( say, $a$) and power of $x$ in each term of  expansion represents sum of value of all four variables. Note that in each term of expansion, some power of $x$ must come from each of $4$ factors.
We can rewrite this series as:
$$\begin{align}
 (1+x+x^2 + ... x^7)^{4}  &=  \left(\frac{1-x^8}{1-x}\right)^4 \\
 &= (1-x^8)^4(1-x)^{-4} \\
 &= \left(\binom{4}{0}-\binom{4}{1}x^8 + \binom{4}{2}x^{16} ...\right) \left(1-x\right)^{-4}
\end{align}$$
Thus for $x^{19}$ we only need coefficients of $x^{19}$, $x^{11}$ and $x^{3}$ from second factor, ie,$(1-x)^{-4}$. Now coefficient of $x^{r}$ in $(1-x)^{-n}$ is $\binom{n+r-1}{r}$, so we have the following answer:
$$\binom{4}{0} \binom{22}{3}-\binom{4}{1} \binom{14}{3}+\binom{4}{2} \binom{6}{3}$$
Which matches with Mike's answer found using Principal of Inclusion Exclusion.
