If `x+y+z=10`, where `x,y,z in N`. Find the number of solutions of triplet (x, y, z) such that no ttwo variables are equal.

Answer

To solve the problem of finding the number of solutions for the triplet (x, y, z) such that \(x + y + z = 10\) with the conditions that \(x, y, z \in \mathbb{N}\) (natural numbers) and no two variables are equal, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
We need to find the number of ordered triplets (x, y, z) such that:
- \(x + y + z = 10\)
- \(x, y, z\) are natural numbers (i.e., \(x, y, z \geq 1\))
- \(x, y, z\) are all different.

2. **Rearranging the Equation:**
Since \(x, y, z\) are natural numbers, we can express them as:
- Let \(x' = x - 1\)
- Let \(y' = y - 1\)
- Let \(z' = z - 1\)

This transformation gives us \(x', y', z' \geq 0\) (non-negative integers). The equation becomes:
\[ (x' + 1) + (y' + 1) + (z' + 1) = 10 \]
Simplifying, we have:
\[ x' + y' + z' = 7 \]

3. **Finding Possible Values for z:**
We can choose values for \(z\) from 1 to 8 (since \(z\) must be less than 10). For each value of \(z\), we will find the corresponding values of \(x\) and \(y\).

4. **Calculating Solutions for Each z:**
- For each value of \(z\), we calculate \(x + y = 10 - z\).
- We will then find pairs \((x, y)\) such that \(x\) and \(y\) are both natural numbers and different from \(z\).

5. **Counting Valid Combinations:**
- For \(z = 1\): \(x + y = 9\) → Valid pairs: (2, 7), (3, 6), (4, 5) → 3 solutions.
- For \(z = 2\): \(x + y = 8\) → Valid pairs: (1, 7), (3, 5), (4, 6) → 3 solutions.
- For \(z = 3\): \(x + y = 7\) → Valid pairs: (1, 6), (2, 5), (4, 3) → 3 solutions.
- For \(z = 4\): \(x + y = 6\) → Valid pairs: (1, 5), (2, 4), (3, 3) → 2 solutions (3, 3 not valid).
- For \(z = 5\): \(x + y = 5\) → Valid pairs: (1, 4), (2, 3) → 2 solutions.
- For \(z = 6\): \(x + y = 4\) → Valid pairs: (1, 3) → 1 solution.
- For \(z = 7\): \(x + y = 3\) → Valid pairs: (1, 2) → 1 solution.
- For \(z = 8\): \(x + y = 2\) → No valid pairs.

6. **Totaling the Solutions:**
Now, we sum the valid combinations:
- For \(z = 1\): 3 solutions
- For \(z = 2\): 3 solutions
- For \(z = 3\): 3 solutions
- For \(z = 4\): 2 solutions
- For \(z = 5\): 2 solutions
- For \(z = 6\): 1 solution
- For \(z = 7\): 1 solution
- For \(z = 8\): 0 solutions

Total = \(3 + 3 + 3 + 2 + 2 + 1 + 1 + 0 = 15\) solutions.

### Final Answer:
The total number of solutions for the triplet (x, y, z) such that \(x + y + z = 10\) with \(x, y, z \in \mathbb{N}\) and no two variables equal is **15**.