Question

The total number of numbers greater than 1000 but less than 4000 that can be formed using 0, 2, 3, 4 (using repetition allowed) are

• A. 125
• B. 105
• C. 128
• D. 625

Hint:

When forming numbers with repetition allowed, multiply the choices for each digit position. Consider restrictions on the first digit when necessary.

Answer 1

The Correct Option is C

To solve this problem, we need to determine how many numbers greater than 1000 and less than 4000 can be formed using the digits 0, 2, 3, and 4, with repetition allowed. These numbers are four-digit numbers, and the thousands digit determines if the number is between 1000 and 4000.

• Thousands Digit: The valid digits for the thousands place, ensuring the number is greater than 1000 but less than 4000, are 1, 2, and 3. However, 1 is not an available digit. Hence, the choices for the thousands digit are 2, 3.
• Hundreds, Tens, and Units Digits: For these positions, we can use any digit from the set {0, 2, 3, 4}.

For each specific thousands digit, the number of possible combinations for the hundreds, tens, and units digits is calculated as follows:

• Each of these three positions has 4 possible digits (0, 2, 3, 4).
• Thus, for each thousands digit, the total number of combinations is: \(4 \times 4 \times 4 = 4^3 = 64\).

Now, compute the total number of numbers:

• For Thousands Digit "2":
  • Number of numbers is 64.
• For Thousands Digit "3":
  • Number of numbers is 64.

Adding these possibilities gives:

\(64 + 64 = 128\)

Therefore, the total number of numbers greater than 1000 but less than 4000 that can be formed is 128.

Answer 2

To find the total number of numbers greater than 1000 but less than 4000 that can be formed using the digits 0, 2, 3, and 4 with repetition allowed, we proceed as follows:

1. We are forming numbers with 4 digits: The number is represented as "abcd," where each letter represents a digit.

2. The first digit, 'a', can only be 1, 2, or 3 to ensure the number is greater than or equal to 1000 and less than 4000. Since the given digits are 0, 2, 3, and 4:

• 'a' can be either 2 or 3.

3. The digits 'b', 'c', and 'd' can each be 0, 2, 3, or 4.

Calculating the possibilities:

• For 'a': 2 choices (2, 3)
• For 'b': 4 choices (0, 2, 3, 4)
• For 'c': 4 choices (0, 2, 3, 4)
• For 'd': 4 choices (0, 2, 3, 4)

4. Therefore, the total number of numbers = 2 × 4 × 4 × 4 = 2 × 64 = 128.

Thus, there are a total of 128 numbers that can be formed.