If the total number of positive integral solution of `15ltx_(1)+x_(2)+x_(3)le20` is k, then the value of `(k)/(100)` is equal to

Answer

To solve the problem of finding the total number of positive integral solutions of the inequality \(15 < x_1 + x_2 + x_3 \leq 20\), we can break it down into steps.

### Step 1: Understand the inequality
We need to find the number of positive integral solutions for the equation \(x_1 + x_2 + x_3 = n\) where \(n\) can take values from 16 to 20 (inclusive). This is because \(x_1 + x_2 + x_3\) must be greater than 15 and less than or equal to 20.

### Step 2: Set up the equation
We will consider the cases for \(n = 16, 17, 18, 19, 20\).

### Step 3: Use the stars and bars theorem
The number of positive integral solutions of the equation \(x_1 + x_2 + x_3 = n\) can be found using the formula:
\[ \text{Number of solutions} = \binom{n-1}{k-1} \]
where \(k\) is the number of variables (in this case, \(k = 3\)).

### Step 4: Calculate for each case
1. **For \(n = 16\)**:
\[ \text{Number of solutions} = \binom{16-1}{3-1} = \binom{15}{2} = \frac{15 \times 14}{2} = 105 \]

2. **For \(n = 17\)**:
\[ \text{Number of solutions} = \binom{17-1}{3-1} = \binom{16}{2} = \frac{16 \times 15}{2} = 120 \]

3. **For \(n = 18\)**:
\[ \text{Number of solutions} = \binom{18-1}{3-1} = \binom{17}{2} = \frac{17 \times 16}{2} = 136 \]

4. **For \(n = 19\)**:
\[ \text{Number of solutions} = \binom{19-1}{3-1} = \binom{18}{2} = \frac{18 \times 17}{2} = 153 \]

5. **For \(n = 20\)**:
\[ \text{Number of solutions} = \binom{20-1}{3-1} = \binom{19}{2} = \frac{19 \times 18}{2} = 171 \]

### Step 5: Sum the solutions
Now, we sum the solutions for \(n = 16\) to \(n = 20\):
\[ k = 105 + 120 + 136 + 153 + 171 = 685 \]

### Step 6: Calculate \(\frac{k}{100}\)
Finally, we need to find the value of \(\frac{k}{100}\):
\[ \frac{k}{100} = \frac{685}{100} = 6.85 \]

### Final Answer
Thus, the value of \(\frac{k}{100}\) is \(6.85\).

---