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 $a , b> 1$ 

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 $$\frac{(n+3)^2}{12}$$
