What is the number of odd integers between 1000 and 9999 with no digit repeated ?

Options

A. 2100
B. 2120
C. 2240
D. 3331

Explanation

To find the number of odd integers between 1000 and 9999 with no digit repeated, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Range**: We are looking for 4-digit odd integers between 1000 and 9999. This means the first digit must be from 1 to 9 (it cannot be 0), and the last digit must be odd (1, 3, 5, 7, or 9).

2. **Choose the Last Digit**: The last digit must be an odd number. The possible choices for the last digit are 1, 3, 5, 7, or 9. This gives us 5 options.

3. **Choose the First Digit**: The first digit can be any digit from 1 to 9, but it cannot be the same as the last digit. Therefore, if we have already chosen the last digit, we have 8 options left for the first digit.

4. **Choose the Middle Digits**: After selecting the first and last digits, we need to fill the second and third digits. The digits can be any digit from 0 to 9, excluding the digits already chosen for the first and last positions. This leaves us with 8 remaining digits (10 total digits - 2 already used).

5. **Calculate the Combinations**:
- The number of ways to choose the last digit: **5** (1, 3, 5, 7, or 9).
- The number of ways to choose the first digit after selecting the last digit: **8**.
- The number of ways to choose the second digit: **8** (from the remaining digits).
- The number of ways to choose the third digit: **7** (from the remaining digits after choosing the first, last, and second).

6. **Total Combinations**: Multiply the number of choices together:
\[ \text{Total} = (\text{choices for last}) \times (\text{choices for first}) \times (\text{choices for second}) \times (\text{choices for third}) \]
\[ \text{Total} = 5 \times 8 \times 8 \times 7 \]

7. **Calculate the Final Answer**:
\[ \text{Total} = 5 \times 8 \times 8 \times 7 = 5 \times 448 = 2240 \]

Thus, the total number of odd integers between 1000 and 9999 with no digit repeated is **2240**.