What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated?

Correct Option (A)

To form 4-digit numbers less than 2000 using the digits 1, 2, 3, and 4 without repetition, the thousands digit must be 1. The remaining three digits (2, 3, and 4) are then arranged in the hundreds, tens, and units places.

• The number of permutations for the remaining three digits is 3! = 3 × 2 × 1 = 6. Therefore, there are 6 such 4-digit numbers.
• The sum of these numbers is calculated by summing the contributions from each place value:
  • Thousands place: The digit 1 occupies the thousands place for all 6 numbers.
    Contribution = 1 × 1000 × 6 = 6000.
  • Hundreds, Tens, and Units places: For these positions, the digits 2, 3, and 4 are permuted. Each of these three digits appears an equal number of times in each of these three positions. The frequency of each digit in any of these places is 3! / 3 = 2 times.
    The sum of these digits is 2 + 3 + 4 = 9.
    • Hundreds place contribution = 100 × (2 + 3 + 4) × 2 = 100 × 9 × 2 = 1800.
    • Tens place contribution = 10 × (2 + 3 + 4) × 2 = 10 × 9 × 2 = 180.
    • Units place contribution = 1 × (2 + 3 + 4) × 2 = 1 × 9 × 2 = 18.
• Total sum = 6000 + 1800 + 180 + 18 = 7998.

Incorrect Options:

Options B, C, and D are incorrect as the systematic calculation of the sum of all 4-digit numbers less than 2000, formed by the digits 1, 2, 3, and 4 without repetition, yields 7998.