The number of three digit numbers of the form xyz such that `x lt y , z le y and x ne0`, is

Options

A. 240
B. 244
C. 276
D. 285

Explanation

To solve the problem of finding the number of three-digit numbers of the form xyz such that \( x < y \), \( z \leq y \), and \( x \neq 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Range for y**:
- Since \( x \) cannot be 0 (it is a three-digit number), \( y \) must be at least 2. Therefore, the possible values for \( y \) are from 2 to 9.

2. **Determine the Possible Values for x**:
- For each value of \( y \), \( x \) can take values from 1 to \( y-1 \) (since \( x < y \)).
- Therefore, the number of choices for \( x \) when \( y = n \) (where \( n \) is the current value of \( y \)) is \( n - 1 \).

3. **Determine the Possible Values for z**:
- The condition \( z \leq y \) means that \( z \) can take values from 0 to \( y \).
- Therefore, the number of choices for \( z \) when \( y = n \) is \( n + 1 \) (since \( z \) can be 0, 1, 2, ..., up to \( n \)).

4. **Calculate the Total Combinations**:
- For each value of \( y \) from 2 to 9, the total combinations of \( x \) and \( z \) can be calculated as:
\[ \text{Total combinations for a specific } y = (n - 1) \times (n + 1) \]
- We will sum this for \( n \) from 2 to 9.

5. **Summation**:
- The total number of three-digit numbers can be expressed as:
\[ \text{Total} = \sum_{n=2}^{9} (n - 1)(n + 1) \]
- Simplifying \( (n - 1)(n + 1) \) gives \( n^2 - 1 \). Thus:
\[ \text{Total} = \sum_{n=2}^{9} (n^2 - 1) \]
- This can be separated into two sums:
\[ \text{Total} = \sum_{n=2}^{9} n^2 - \sum_{n=2}^{9} 1 \]

6. **Calculating the Sums**:
- The sum of squares from 1 to \( n \) is given by the formula:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
- For \( n = 9 \):
\[ \sum_{k=1}^{9} k^2 = \frac{9 \times 10 \times 19}{6} = 285 \]
- For \( n = 1 \):
\[ \sum_{k=1}^{1} k^2 = 1 \]
- Thus:
\[ \sum_{n=2}^{9} n^2 = 285 - 1 = 284 \]
- The count of terms from 2 to 9 is \( 8 \), so:
\[ \sum_{n=2}^{9} 1 = 8 \]

7. **Final Calculation**:
- Therefore:
\[ \text{Total} = 284 - 8 = 276 \]

### Conclusion:
The total number of three-digit numbers of the form xyz such that \( x < y \), \( z \leq y \), and \( x \neq 0 \) is **276**.