Find the number of solutions of the equation `x+y+z=6`, where `x, y, z in W`

Answer

To find the number of solutions of the equation \(x + y + z = 6\) where \(x, y, z \in W\) (the set of whole numbers), we can use the combinatorial method known as "stars and bars." Here’s a step-by-step solution:

### Step 1: Understanding the Problem
We need to find the number of non-negative integer solutions to the equation \(x + y + z = 6\). This means \(x\), \(y\), and \(z\) can take values from the set of whole numbers (0, 1, 2, ...).

### Step 2: Applying the Stars and Bars Theorem
According to the stars and bars theorem, the number of ways to distribute \(n\) indistinguishable objects (stars) into \(k\) distinguishable boxes (variables) is given by the formula:

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

In our case:
- \(n = 6\) (the total we want to achieve)
- \(k = 3\) (the number of variables: \(x\), \(y\), and \(z\))

### Step 3: Plugging in the Values
Using the formula, we substitute \(n\) and \(k\):

\[ \text{Number of solutions} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} \]

### Step 4: Calculating the Binomial Coefficient
Now we calculate \(\binom{8}{2}\):

\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 \]

### Conclusion
Thus, the number of non-negative integer solutions to the equation \(x + y + z = 6\) is \(28\).

---