If the number of solutions of the equation `x+y+z=20,` where `1 le x lt y lt z` and `x, y, z in I` is k, then `(k)/(10)` is equal to

Answer

To solve the problem of finding the number of integer solutions to the equation \( x + y + z = 20 \) under the constraints \( 1 \leq x < y < z \) and \( x, y, z \in \mathbb{I} \) (the set of positive integers), we can follow these steps:

### Step 1: Understand the constraints
We have the equation \( x + y + z = 20 \) and the conditions \( 1 \leq x < y < z \). This means that \( x \) must be at least 1, and \( y \) and \( z \) must be greater than \( x \).

### Step 2: Substitute values for \( x \)
We will consider different values for \( x \) starting from 1 and find valid pairs \( (y, z) \) for each case.

### Step 3: Case 1: \( x = 1 \)
If \( x = 1 \):
\[ y + z = 20 - 1 = 19 \]
Since \( y < z \), the smallest value \( y \) can take is 2. Thus, we can have:
- \( y = 2, z = 17 \)
- \( y = 3, z = 16 \)
- \( y = 4, z = 15 \)
- \( y = 5, z = 14 \)
- \( y = 6, z = 13 \)
- \( y = 7, z = 12 \)
- \( y = 8, z = 11 \)
- \( y = 9, z = 10 \)

This gives us **8 solutions**.

### Step 4: Case 2: \( x = 2 \)
If \( x = 2 \):
\[ y + z = 20 - 2 = 18 \]
The smallest value \( y \) can take is 3. Thus, we can have:
- \( y = 3, z = 15 \)
- \( y = 4, z = 14 \)
- \( y = 5, z = 13 \)
- \( y = 6, z = 12 \)
- \( y = 7, z = 11 \)
- \( y = 8, z = 10 \)

This gives us **6 solutions**.

### Step 5: Case 3: \( x = 3 \)
If \( x = 3 \):
\[ y + z = 20 - 3 = 17 \]
The smallest value \( y \) can take is 4. Thus, we can have:
- \( y = 4, z = 13 \)
- \( y = 5, z = 12 \)
- \( y = 6, z = 11 \)
- \( y = 7, z = 10 \)
- \( y = 8, z = 9 \)

This gives us **5 solutions**.

### Step 6: Case 4: \( x = 4 \)
If \( x = 4 \):
\[ y + z = 20 - 4 = 16 \]
The smallest value \( y \) can take is 5. Thus, we can have:
- \( y = 5, z = 11 \)
- \( y = 6, z = 10 \)
- \( y = 7, z = 9 \)

This gives us **3 solutions**.

### Step 7: Case 5: \( x = 5 \)
If \( x = 5 \):
\[ y + z = 20 - 5 = 15 \]
The smallest value \( y \) can take is 6. Thus, we can have:
- \( y = 6, z = 9 \)
- \( y = 7, z = 8 \)

This gives us **2 solutions**.

### Step 8: Case 6: \( x = 6 \)
If \( x = 6 \):
\[ y + z = 20 - 6 = 14 \]
The smallest value \( y \) can take is 7. However, \( y \) cannot be less than or equal to \( z \) in this case, so there are **no solutions**.

### Step 9: Total Solutions
Now, we sum the solutions from all cases:
- Case 1: 8 solutions
- Case 2: 6 solutions
- Case 3: 5 solutions
- Case 4: 3 solutions
- Case 5: 2 solutions
- Case 6: 0 solutions

Total solutions \( k = 8 + 6 + 5 + 3 + 2 + 0 = 24 \).

### Step 10: Calculate \( \frac{k}{10} \)
Now, we find \( \frac{k}{10} \):
\[ \frac{24}{10} = 2.4 \]

### Final Answer
Thus, the value of \( \frac{k}{10} \) is **2.4**.