How many positive integers solutions does $x + y + z + w = 16$ have if $x < y$

Question

$x + y + z + w = 16$ is relatively easy, but I'm stuck in the restriction $x < y$.

Answer 1

Solution sketch: All the solutions (and you already know how many) are divided into three types: $x<y$, $x = y$ or $x>y$.

The number of solutions with $x = y$, I'm confident you can find.

Exactly half of the remaining solutions have $x<y$.

---

Edit: The $x = y$ case.

For this, we are looking for the number of solutions to $$ 2x + z + w = 16 $$ I'm tempted to just to a brute force solution here, as there are so few cases: $$ \begin{array}{|c|cccccccc|} \hline x & 1&2&3&4&5&6&7\\ \text{Number of solutions}&13&11&9&7&5&3&1\\ \hline \end{array} $$ And the sum of all of these is $7^2 = 49$ (using the well-known fact that the sum of consecutive odd numbers starting at $1$ is a perfect square). But even if you didn't know this, it isn't difficult to add up: $$ (13+1) + (1+3) + (9+5) + 7 = 14+14+14+7 = 3.5\cdot 14 = 7\cdot 7 $$

Answer 2

If $x<y$, then for some positive $k \in \mathbb{Z}$, $x+k=y$. By substitution,

$$2x+k+z+w=16$$