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 $a , b> 1$ 

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 $f$:
$$f(t)=\frac{(1-t)^{-3}}{6}+\frac{(1+t)\,\left(1-t^2\right)^{-2}}{2}+\frac{\left(1-t^3\right)^{-1}}{3}\,.$$
