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

Options

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

Answer

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

### Step 1: Identify the range and properties of the number
We are looking for 4-digit odd integers that range from 1000 to 9999. Since we want odd integers, the last digit (the units place) must be an odd digit.

### Step 2: Determine the possible odd digits
The odd digits available are 1, 3, 5, 7, and 9. This gives us a total of 5 choices for the last digit.

### Step 3: Choose the last digit
Let's denote the last digit as \(d_4\). Since \(d_4\) must be odd, we can choose it from the set {1, 3, 5, 7, 9}.

### Step 4: Choose the first digit
The first digit (thousands place, \(d_1\)) cannot be 0 and cannot be the same as the last digit. Therefore, if we have already chosen one of the 5 odd digits for \(d_4\), we have 8 remaining choices for \(d_1\) (0-9 excluding the digit chosen for \(d_4\) and also excluding 0).

### Step 5: Choose the second digit
The second digit (hundreds place, \(d_2\)) can be any digit from 0-9, excluding the digits already chosen for \(d_1\) and \(d_4\). This gives us 8 remaining choices for \(d_2\).

### Step 6: Choose the third digit
The third digit (tens place, \(d_3\)) can also be any digit from 0-9, excluding the digits already chosen for \(d_1\), \(d_2\), and \(d_4\). This gives us 7 remaining choices for \(d_3\).

### Step 7: Calculate the total number of combinations
Now, we can calculate the total number of odd integers as follows:

\[ \text{Total combinations} = \text{Choices for } d_4 \times \text{Choices for } d_1 \times \text{Choices for } d_2 \times \text{Choices for } d_3 \]
\[ = 5 \times 8 \times 8 \times 7 \]
\[ = 5 \times 8 \times 8 \times 7 = 2240 \]

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