Extended Stars-and-Bars where the upper bound of a variable is one of the other variables How many non-negative integer solutions are there for the equation x + y + z + w = 20, where x>y.
I started by substituting x for a new variable x'=x-y:
x > y
x-y > 0
x'>0  
However, this led me to a point from which I didn't know how to make progress:
x'+ 2y + z + w = 20
How should I deal with the 2y? 
(the answer provided was 825)
 A: By symmetry, the number of solutions with $x>y$ is the same as the number with $x<y$ so the answer is "half the number of solutions where $x\neq y$."  I would go about this by subtracting the solutions with $x=y$ from the total number of solutions.
Stars and bars gives the total solutions, so now we need to know the number of solutions to $$2x+w+z=20$$ in nonnegative integers.  For a given $x$ this is just $21-2x$ so it's easy to add them up. 
A: $$x+y+z+w=20\tag{1}$$
Firstly it says $x \gt y$ so $x=y+x'+1$ where $x'\ge 0$. 
This gives
$$x'+2y+z+w=19\, .$$
So since $2y$ is always even, $x'+z+w$ must be odd which means either all three variables are odd or just one is.


*

*If all three are odd then say $x'=2x''+1$, $z=2z''+1$, $w=2w''+1$, giving


\begin{align}&& 2(x''+y+z''+w'')&=16\\[1ex]
&\implies& x''+y+z''+w''&=8\, ,\end{align}
which has $\binom{8+3}{3}$ non-negative integer solutions by bars and stars.


*If just one is odd, say $x'=2x''+1$, then the other two must be even: $z=2z''$, $w=2w''$, this gives


\begin{align}&& 2(x''+y+z''+w'')&=18\\[1ex]
&\implies & x''+y+z''+w''&=9\, ,\end{align}
which has $\binom{9+3}{3}$ non-negative integer solutions by bars and stars. However, there are 3 choices for our odd variable so that's $3\binom{9+3}{3}$ total solutions for this case.
Adding both cases together we have a grand total of
$$\binom{8+3}{3}+3\binom{9+3}{3}=825\tag{Answer}$$
non-negative integer solutions to $(1)$ with $x\gt y$.
