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 $a , b> 1$ 

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
$$-\frac{1}{9} + \frac{1}{8} + \frac{17}{72} + \frac{1}{4} \binom{201}{1} + \frac{1}{6} \binom{202}{2} = 3434.$$
