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!
 A: In the second case you can write $x_1=2k, x_2=2m+1, x_3=2n+1.$ Then 
$$2018=x_1+x_2+x_3=2k+2m+1+2n+1,$$ what gives "stars and bars" for equation $1008=k+m+n$.
Ps. Always specify the domains for $x$'s - are they positive, non-negative or arbitrary integers? All these three variants of "stars and bars" have different solutions.
A: Use generating functions to solve the the second question. Indeed,we seek the coefficient of $x^{2018}$ in the series expansion of
$$
(x^0+x^2+x^4+\dotsb)(x+x^3+x^5+\dotsb)^2=\frac{1}{1-x^2}\times\frac{x^2}{(1-x^2)^2}=\frac{x^2}{(1-x^2)^3}
$$
Thus we seek the coefficient of $x^{2016}$ in the series expansion of
$$
(1-x^2)^{-3}=\sum_{n=0}^\infty\binom{n+2}{2}x^{2n}
$$
Hence the number of solutions is
$$
\binom{1008+2}{2}=\binom{1010}{2}.
$$
A: Hint: Even numbers can be uniquely written in the form $2k$ and odd numbers can be uniquely written in the form $2k+1$ for some integer $k$.
A: Indeed stars and bars, and your idea about 1) is okay.
Hint on 2):
Find the number of solutions of $y_1+y_2+y_3=1008$.
If $(y_1,y_2,y_3)$ is such a solution then $(2y_1,2y_2+1,2y_2+1)$ will  satisfy the conditions under 2).
A: 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 $$\sum_{k=0}^{1009}{(1009-k)}=\sum_{k=0}^{1009}{k}=\frac{1009\times1010}{2}=509545.$$
