<s>[INST] How many are the non-negative integer solutions of $ +  +  +  = 16$ where $x < y$? 
How many are the non-negative integer solutions of $ +  +  +  = 16$ where $x < y$?

Anyone can explain how to think to approach this type of problem? 
The answer is 444.
 [/INST] Consider cases and find a pattern. 
When $x=0$:
$$0+y+z+w=16, y\ge 1, z,w\ge 0 \Rightarrow \\
t+1+z+w=16,t,z,w\ge 0 \Rightarrow \\
t+z+w=15, t,z,w\ge 0 \Rightarrow \\
{15+3-1\choose 3-1}$$ 
Note: It was used Stars and Bars.
When $x=1$:
$$1+y+z+w=16, y\ge 2, z,w\ge 0 \Rightarrow \\
t+2+z+w=15,t,z,w\ge 0 \Rightarrow \\
t+z+w=13, t,z,w\ge 0 \Rightarrow \\
{13+3-1\choose 3-1}$$
$\vdots$
When $x=7$:
$$7+y+z+w=16, y\ge 8, z,w\ge 0 \Rightarrow \\
t+8+z+w=9,t,z,w\ge 0 \Rightarrow \\
t+z+w=1, t,z,w\ge 0 \Rightarrow \\
{1+3-1\choose 3-1}$$ 
Hence, it is the sum:
$$\begin{align}\sum_{i=1}^8 {2i-1+3-1\choose 3-1}&=\sum_{i=1}^8 {2i+1\choose 2}=\sum_{i=1}^8 \frac{(2i+1)!}{2!(2i-1)!}=\sum_{i=1}^8 \frac{(2i+1)2i}{2}=\\
&=\sum_{i=1}^8 (2i^2+i)=2\frac{8(8+1)(2\cdot 8+1)}{6}+\frac{8(8+1)}{2}=\\
&=24\cdot 17+36=444.\end{align}$$
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