Find no. of non negative integer solutions of $a+2b+3c = 200$
Using generating functions, this would be the coefficient of $x^{200}$ in $$f(x) = \frac{1}{1-x} \cdot \frac{1}{1-x^2} \cdot \frac{1}{1-x^3} = \frac{1}{(1-x)^3 (1+x) (1+x+x^2)}.$$ Now, according to Maxima, if you calculate a partial fractions expansion of $f(x)$ you should get: $$f(x) = \frac{2+x}{9(1+x+x^2)} + \frac{1}{8(1+x)} + \frac{17}{72(1-x)} + \frac{1}{4(1-x)^2} + \frac{1}{6(1-x)^3} = \\ \frac{2-x-x^2}{9(1-x^3)} + \frac{1}{8(1+x)} + \frac{17}{72(1-x)} + \frac{1}{4(1-x)^2} + \frac{1}{6(1-x)^3}.$$ Therefore, the coefficient of $x^{200}$ would be $$-\frac{1}{9} + \frac{1}{8} + \frac{17}{72} + \frac{1}{4} \binom{201}{1} + \frac{1}{6} \binom{202}{2} = 3434.$$
You can't use stars and bars since $2b$ and $3c$ cannot be any integer. However, a nice way to do this generally for $a+2b+3c=n$ is to make the substitution $x=c$, $y=b+c$ and $z=a+b+c$ so we have $x \leq y \leq z$. Then it comes down to some casework for when all three are different, two are the same, and all three are the same.
The total amount of cases without $x \leq y \leq z$ would be $\binom{n+2}{2}$ by stars and bars, and this will help us in that we only need to consider two of the cases, then solve for the last.
Case 1: All three are the same. For $n=200$, there will not be a case when all three are the same since $3$ does not divide $200$.
Case 2: Two numbers are the same. After beginning to list out the possibilities $$x=n \quad \quad y,z=0$$ $$x=n-2 \quad \quad y,z=1$$ $$\cdots$$ $$x=n-2\left\lfloor \frac{n}{2} \right\rfloor \quad \quad y,z=\left\lfloor \frac{n}{2} \right\rfloor$$ It is clear that there are $\left\lfloor \frac{n}{2} \right\rfloor +1$ cases. And we would have counted each one three times in our $\binom{n+2}{2}$ since there we do not include the condition $x \leq y \leq z$.
Case 3: All three are different. We can now calculate these using our total $\binom{n+2}{2}$ from before to get that there are $$\binom{n+2}{2}-3\left(\left\lfloor \frac{n}{2} \right\rfloor+1\right)$$ solutions for $x,y,z$ without $x \leq y \leq z$ and thus $$\frac{\binom{n+2}{2}-3\left(\left\lfloor \frac{n}{2} \right\rfloor+1\right)}{6}$$ solutions with $x \leq y \leq z$
Now we can finally add our cases up to get $$\frac{\binom{n+2}{2}-3\left(\left\lfloor \frac{n}{2} \right\rfloor+1\right)}{6}+\left(\left\lfloor \frac{n}{2} \right\rfloor+1\right)$$ $$\frac{\binom{n+2}{2}+3\left(\left\lfloor \frac{n}{2} \right\rfloor+1\right)}{6}$$
And for $n=200$, we get the answer $3434$.
Addendum: I like to solve things generally, so here it is for numbers that are divisible by $3$. We'll have $1$ solution for our Case 1 and have to subtract off $1$ from Case 2 since there'll be a solution $x=y=z$. So, instead we'd have $$\frac{\binom{n+2}{2}-3\left\lfloor \frac{n}{2} \right\rfloor - 1}{6} + \left\lfloor \frac{n}{2} \right\rfloor + 1$$ or $$\frac{\binom{n+2}{2}+3\left\lfloor \frac{n}{2} \right\rfloor +5}{6}$$
Also, it's fun to notice that this is the closest integer to $$\frac{(n+3)^2}{12}$$
Outline:
Step 1: How many solutions are there to $a+2b=n$ where $n$ is a fixed nonnegative integer. Answer: $b$ can be any of $0, 1, 2,..., \left\lfloor \frac{n}{2} \right\rfloor$. So there are $\left\lfloor \frac{n}{2} \right\rfloor + 1$ solutions.
Step 2: $n$ (from step 1) can be anything of the form $200 - 3c$ where $c$ takes on values from $0$ to $66$. [That is the values of $n$ are $200, 197, 194, ...,2$.]
Step 3: For each value of $n$ in step 2 you will get some number of solutions (based on step 1). Add these all together. There may be helpful patterns you can exploit.