The number of three digit numbers of the form xyz such that `x lt y , z le y and x ne0`, is a ) 240 b) 244 c) 276 d) 285

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 break down the solution into several steps.

### Step-by-Step Solution:

1. **Identify the Range of Digits**:
- The digits for \( x, y, z \) can range from 0 to 9, but since \( x \) cannot be 0 (as it is the leading digit of a three-digit number), \( x \) can only take values from 1 to 9. Therefore, we have 9 possible choices for \( x \) (1 to 9).

2. **Conditions on Digits**:
- The conditions given are:
- \( x < y \)
- \( z \leq y \)
- This means \( y \) must be greater than \( x \), and \( z \) can be equal to or less than \( y \).

3. **Choosing Digits**:
- We can choose \( y \) from the digits greater than \( x \). If \( x = 1 \), \( y \) can be from 2 to 9 (8 options). If \( x = 2 \), \( y \) can be from 3 to 9 (7 options), and so on until \( x = 8 \) where \( y \) can only be 9 (1 option). Therefore, the number of choices for \( y \) based on \( x \) can be summarized as follows:
- For \( x = 1 \): 8 choices for \( y \)
- For \( x = 2 \): 7 choices for \( y \)
- For \( x = 3 \): 6 choices for \( y \)
- For \( x = 4 \): 5 choices for \( y \)
- For \( x = 5 \): 4 choices for \( y \)
- For \( x = 6 \): 3 choices for \( y \)
- For \( x = 7 \): 2 choices for \( y \)
- For \( x = 8 \): 1 choice for \( y \)

4. **Counting Valid Combinations**:
- For each valid pair of \( (x, y) \), we need to determine the number of valid values for \( z \).
- Since \( z \) can take any value from 0 to \( y \) (inclusive), the number of choices for \( z \) is \( y + 1 \) (including 0).

5. **Calculate Total Combinations**:
- We can now calculate the total number of valid combinations:
- For \( x = 1 \): \( y = 2 \) to \( 9 \) gives \( 8 + 1, 7 + 1, 6 + 1, 5 + 1, 4 + 1, 3 + 1, 2 + 1, 1 + 1 \) choices for \( z \).
- The total combinations can be calculated as:
\[ \text{Total} = \sum_{x=1}^{8} \text{(choices for } y) \times (y + 1) \]
- This results in:
\[ = 8 \times 9 + 7 \times 8 + 6 \times 7 + 5 \times 6 + 4 \times 5 + 3 \times 4 + 2 \times 3 + 1 \times 2 \]
- Which simplifies to:
\[ = 72 + 56 + 42 + 30 + 20 + 12 + 6 + 2 = 240 \]

6. **Final Count**:
- After calculating all combinations, we find that the total number of three-digit numbers of the form \( xyz \) satisfying the conditions is **240**.

### Final Answer:
Thus, the answer is **a) 240**.