Find no. of non negative integer solutions of $a+2b+3c = 200$ I want to find number of the non negative integer solutions of $a+2b+3c = 200$
(answer can contain $a,b,c = 0$)
Example one of the solutions is $a = 100, b = 50, c = 0$
I used stars and bars trick and here i get is
$\dfrac{\dbinom{200+2}{2}}{2 \times 3}$ but it is incorrect
any type of math is welcomed

what I have learnt from stars and bars technique is to find the solutions of $x+y = n , x,y,n > 0$ are positive integers but I am confused with equation of type $ax+by = n$ where $a , b> 1$ 
 A: 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 $$\frac{(n+3)^2}{12}$$
A: 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
$$-\frac{1}{9} + \frac{1}{8} + \frac{17}{72} + \frac{1}{4} \binom{201}{1} + \frac{1}{6} \binom{202}{2} = 3434.$$
A: 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.
A: Consider $x:=a+b+c$, $y:=b+c$, and $z:=c$.  Let $n:=200$.  Then, $x\geq y\geq z\geq 0$ and $x+y+z=n$.  Let $$T_n:=\big\{(x,y,z)\in\mathbb{Z}_{\geq 0}^3\,|\,x+y+z=n\big\}\,.$$  Then, the symmetric group $S_3$ of order $3!=6$ acts on $T_n$ by permuting the three entries of each element of $T_n$.  The number of $S_3$-orbits $N_n$ in $T_n$ is precisely the number of triples $(x,y,z)\in T_n$ with $x\geq y\geq z$.
By Burnside's Lemma,
$$N_n=\frac{1}{|S_3|}\,\sum_{g\in S_3}\,\big|\text{Fix}(g)\big|\,,$$
where $\text{Fix}(g)$ denotes the number of triples $(x,y,z)\in T_n$ fixed by $g$.  The class equation of $S_3$ is $$\#\{\text{identity}\}+\#\{\text{transpositions}\}+\#\{\text{$3$-cycles}\}=1+3+2\,.$$
That is,
$$N_n=\frac{1}{6}\,\Biggl(1\cdot \binom{n+3-1}{3-1}+3\cdot\left\lfloor \frac{n+2}{2}\right\rfloor+2\cdot \left\lfloor\frac{3-(n\text{ mod }3)}{3}\right\rfloor\Biggr)\,,$$
so
$$N_n=\frac{1}{6}\,\binom{n+2}{2}+\frac{1}{2}\,\Biggl(1+\left\lfloor\frac{n}{2}\right\rfloor\Biggr)+\frac{1}{3}\,\left\lfloor\frac{3-(n\text{ mod }3)}{3}\right\rfloor\,.$$
In particular,
$$N_{200}=\frac{1}{6}\,\binom{202}{2}+\frac{1}{2}\,(1+100)+\frac{1}{3}\,(0)=3434\,.$$

From Daniel Schepler's solution, the generating function $f(t):=\sum\limits_{n=0}^\infty\,N_n\,t^n$ is given by $f(t)=\dfrac{1}{(1-t)\,(1-t^2)\,(1-t^3)}$.  This answer gives a slightly more aesthetic expression for $f$:
$$f(t)=\frac{(1-t)^{-3}}{6}+\frac{(1+t)\,\left(1-t^2\right)^{-2}}{2}+\frac{\left(1-t^3\right)^{-1}}{3}\,.$$
