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?

Here's a brief explanation: 1. Identify the numbers: The numbers must be 4-digit, less than 2000, and formed by digits 1, 2, 3, 4 without repetition. This means the thousands digit must be 1. The remaining digits (2, 3, 4) can be arranged in the hundreds, tens, and units places. There are 3! = 6 such numbers. 2. List the numbers: 1234, 1243, 1324, 1342, 1423, 1432. 3. Calculate the sum by place value: * Thousands place: All 6 numbers have 1 in the thousands place. Sum = 6 * 1000 = 6000. * Hundreds place: The digits 2, 3, 4 each appear 2 times (3! / 3 = 2 times) in the hundreds place. Sum = (2+3+4) * 2 * 100 = 9 * 2 * 100 = 1800. * Tens place: The digits 2, 3, 4 each appear 2 times in the tens place. Sum = (2+3+4) * 2 * 10 = 9 * 2 * 10 = 180. * Units place: The digits 2, 3, 4 each appear 2 times in the units place. Sum = (2+3+4) * 2 * 1 = 9 * 2 * 1 = 18. 4. Total Sum: Add the sums from each place value: 6000 + 1800 + 180 + 18 = 7998. Therefore, option A is correct.