number of solutions to the equation $x_1+x_2+x_3=2018$ with odd\even condition on $x$. I have to find the number of solutions to   $x_1+x_2+x_3=2018$ ,
with the following conditions :


*

*$x_1,x_2,x_3$ are even numbers.

*$x_1$ is even while $x_2$ and $x_3$ are odd.


I guess I have to use stars and bars in order to solve it, so
in case one I reffed to every two stars as one star and then solved this equation  $x_1+x_2+x_3=1009$ (with stars and bars again). but in case two i'm stuck, I don't understand how to divide the stars correctly when odd and even  conditions are involved.
appreciate your help very much!

How many solutions are there for $x_2+x_3=2n$ with both $x_2$ and $x_3$ odd? Well, $x_2$ can range from $1$ to $2n-1$, so there are $n$ solutions. This means that, for a given $x_1=2k$, your equation $x_1+x_2+x_3=2018$ has $\frac{2018-x_1}{2}=1009-k$ solutions. Since $k$ may take values from $0$ to $1009$, the total number of solutions is $$\sum_{k=0}^{1009}{(1009-k)}=\sum_{k=0}^{1009}{k}=\frac{1009\times1010}{2}=509545.$$
