The triple `(x,y,z)` is chosen from the set `{1,2,3,………,n}`, such that `x le y < z`. The number of such triples is

Options

A. `n^(3)`
B. `.^(n)C_(3)`
C. `.^(n)C_(2)`
D. `.^(n)C_(3)+.^(n)C_(2)`

Answer

To solve the problem of finding the number of triples \((x, y, z)\) chosen from the set \(\{1, 2, 3, \ldots, n\}\) such that \(x \leq y < z\), we can break it down into two cases based on the conditions given.

### Step-by-Step Solution:

1. **Understanding the Conditions**:
- We have two conditions for the triples:
1. \(x \leq y < z\)
2. This means that \(x\) can be equal to \(y\) or less than \(y\), but \(y\) must always be less than \(z\).

2. **Case 1: \(x < y < z\)**:
- In this case, \(x\), \(y\), and \(z\) are all distinct.
- We need to choose 3 distinct numbers from the set \(\{1, 2, 3, \ldots, n\}\).
- The number of ways to choose 3 distinct numbers is given by the combination formula \(nC3\).
- Therefore, the number of triples in this case is:
\[ \text{Number of triples} = nC3 \]

3. **Case 2: \(x = y < z\)**:
- Here, \(x\) and \(y\) are the same, and \(z\) is distinct and greater than \(x\).
- We can choose \(x\) (which is equal to \(y\)) from the set, and then choose \(z\) from the remaining numbers that are greater than \(x\).
- If we choose \(x\) to be any number \(k\) (where \(k\) can be from 1 to \(n-1\)), then \(z\) can be chosen from the numbers \((k+1)\) to \(n\).
- The number of choices for \(z\) given \(x = k\) is \(n - k\).
- To find the total number of triples for this case, we can sum up the choices for each possible \(x\):
\[ \text{Number of triples} = \sum_{k=1}^{n-1} (n - k) = (n-1) + (n-2) + \ldots + 1 + 0 = \frac{(n-1)n}{2} \]

4. **Final Calculation**:
- Now we add the results from both cases:
\[ \text{Total number of triples} = nC3 + \frac{(n-1)n}{2} \]

5. **Conclusion**:
- Thus, the total number of triples \((x, y, z)\) such that \(x \leq y < z\) is:
\[ \text{Total} = \binom{n}{3} + \frac{(n-1)n}{2} \]