How many even numbers less than 10,000 can be with the digits 3,5,7,8,9, without any repetition

Options

A. 32
B. 16
C. 44
D. 41

Answer

To find how many even numbers less than 10,000 can be formed using the digits 3, 5, 7, 8, and 9 without repetition, we can break down the problem based on the number of digits in the numbers we can form (1-digit, 2-digit, 3-digit, and 4-digit numbers).

### Step 1: Identify the even digit
The only even digit available from the given set is **8**. Therefore, any even number we form must end with **8**.

### Step 2: Count 1-digit even numbers
The only 1-digit even number we can form is **8** itself.
- **Count**: 1

### Step 3: Count 2-digit even numbers
For a 2-digit number, the last digit must be **8**. The first digit can be any of the remaining digits (3, 5, 7, 9).
- Choices for the first digit: 4 (3, 5, 7, 9)
- **Count**: 4

### Step 4: Count 3-digit even numbers
For a 3-digit number, the last digit must again be **8**. The first two digits can be chosen from the remaining digits (3, 5, 7, 9).
- Choices for the first digit: 4 (3, 5, 7, 9)
- Choices for the second digit: 3 (after choosing the first digit)
- **Count**: 4 (for the first digit) × 3 (for the second digit) = 12

### Step 5: Count 4-digit even numbers
For a 4-digit number, the last digit must be **8**. The first three digits can be chosen from the remaining digits (3, 5, 7, 9).
- Choices for the first digit: 4 (3, 5, 7, 9)
- Choices for the second digit: 3 (after choosing the first digit)
- Choices for the third digit: 2 (after choosing the first and second digits)
- **Count**: 4 (for the first digit) × 3 (for the second digit) × 2 (for the third digit) = 24

### Step 6: Total count of even numbers
Now, we can sum the counts from all the cases:
- 1-digit numbers: 1
- 2-digit numbers: 4
- 3-digit numbers: 12
- 4-digit numbers: 24

**Total = 1 + 4 + 12 + 24 = 41**

### Final Answer
The total number of even numbers less than 10,000 that can be formed with the digits 3, 5, 7, 8, and 9 without repetition is **41**.

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