<s>[INST] 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. 
 [/INST] 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.
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