Question

How many integers, greater than \(999\) but not greater than \(4000\), can be formed with the digits \(0, 1, 2, 3, 4\), if repetition of digits is allowed?

• A. 499
• B. 500
• C. 375
• D. 376
• E. 501

Hint:

When counting numbers with restrictions on the first digit, separate cases based on the highest possible digit.

Answer 1

The Correct Option is C

We consider 4-digit numbers from \(1000\) to \(3999\) and \(4000\) separately. Digits available: \(0, 1, 2, 3, 4\). Case 1: First digit is 1, 2, or 3 (since number \(< 4000\)): - First digit: 3 choices - Remaining 3 digits: each has 5 choices (repetition allowed) Total for this case: \(3 \times 5 \times 5 \times 5 = 375\). Case 2: Number is \(4000\): - First digit = 4, and the rest are 0s → only 1 number. But \(4000\) is allowed, so total = \(375 + 1 = 376\). However, since problem states "not greater than 4000", we include \(4000\) → \(\boxed{376}\).