Question

How many 3 digit even numbers can be formed from the digits (0-9) if repetition of digits are allowed?

• A. 350
• B. 400
• C. 450
• D. 1000

Answer 1

The Correct Option is C

To determine how many 3-digit even numbers can be formed from the digits 0-9 with repetition allowed, follow these steps:

1. Understand that a 3-digit even number can be expressed as \(abc\), where \(a\), \(b\), and \(c\) are digits, and \(c\) is an even digit since the number must be even.
2. Possible even digits for \(c\) are 0, 2, 4, 6, and 8, providing 5 choices for \(c\).
3. The digit \(a\) (hundreds place) must be non-zero because it is a 3-digit number. Thus, it can be one of the digits from 1 to 9, giving us 9 choices for \(a\).
4. The digit \(b\) (tens place) can be any digit from 0 to 9, providing 10 choices for \(b\).
5. Since repetition of digits is allowed, the total number of 3-digit even numbers is calculated by multiplying the number of choices for each digit place: 9 choices for \(a\), 10 choices for \(b\), and 5 choices for \(c\):
   \[9 \times 10 \times 5 = 450\]

Therefore, there are 450 possible 3-digit even numbers formed from the digits 0-9 with repetition allowed.