To find the number of combinations of non-negative integer values for $x$, $y$, and $z$ that satisfy the equation:

$x + y + z = 4$

This is a classic "stars and bars" problem in combinatorics. The formula to find the number of solutions to the equation $x_1 + x_2 + \dots + x_k = n$, where all variables are non-negative integers, is given by:

$\text{Number of solutions} = \binom{n+k-1}{k-1}$

In our case:

• $n = 4$ (the total sum)
• $k = 3$ (the number of variables: $x$, $y$, and $z$)

So the number of combinations is:

$\binom{4+3-1}{3-1} = \binom{6}{2}$

Calculating $\binom{6}{2}$:

$\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$

Thus, there are 15 different combinations of non-negative integers $x$, $y$, and $z$ that satisfy $x + y + z = 4$.

Would you like to explore the specific combinations, or ask for details?

Here are some follow-up questions:

1. What are the specific combinations of values for $x$, $y$, and $z$?
2. How does the "stars and bars" method work in general?
3. Can this method be used if one of the variables must be greater than zero?
4. What if negative integers were allowed as solutions?
5. How would the result change if we increased the sum from 4 to another number?

Tip: The stars and bars method is a powerful tool to count combinations of objects where order doesn't matter and repetition is allowed!