How would you find the number of solutions to $m_1+2m_2+3m_3=n$? Assuming $m_i$ are nonnegative integers
I understand that we need to use $C(n+k-1,n-1)$ here but I am not sure how the coefficients of $m_i$ affect the equation?
For example to find the number of solutions to $m_1+m_2+m_2=n$, we could use the equation above to find that it is $C(n+2,2)=\frac{1}{2}(n+2)(n+1)$ 

We can write a statement that is not easy to calculate and try to simplify:
We know $0\le m_3 \le \left\lfloor \dfrac{n}{3}\right\rfloor$ and once we have chosen $m_3$, $0\le m_2 \le \left\lfloor\dfrac{n-3m_3}{2}\right\rfloor$. And for every choice of $m_3$ and $m_2$, there is exactly one choice of $m_1$, which yields the summation:
$$\sum_{m_3=0}^{\left\lfloor \tfrac{n}{3}\right\rfloor}\sum_{m_2=0}^{\left\lfloor \tfrac{n-3m_3}{2}\right\rfloor} 1 = \sum_{m_3=0}^{\left\lfloor \tfrac{n}{3}\right\rfloor}\left(\left\lfloor\dfrac{n-3m_3}{2}\right\rfloor+1\right)$$
Plugging this into Wolframalpha gives:
$$ = \left\lfloor \dfrac{n}{3} \right\rfloor + \left\lfloor \dfrac{n - 3}{2} \right\rfloor \left(\left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor-1 }{2} \right\rfloor+1\right) + \dfrac{1}{2} \left( -3 \left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor -1 }{2} \right\rfloor^2 - 3 \left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor -1 }{2} \right\rfloor - 3 \left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3}  \right\rfloor }{2}\right\rfloor^2 - 3 \left\lfloor\dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor }{2} \right\rfloor + 2 \left\lfloor\dfrac{n}{2}\right\rfloor \left( \left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3}\right\rfloor }{2}\right\rfloor + 1\right) + 2\right)$$
