Question

The sum of all the 4-digit numbers formed by taking all the digits from 0, 3, 6, 9 without repetition is:

• A. 119592
• B. 115992
• C. 211599
• D. 119952

Hint:

When calculating sums involving permutations of digits in number formation, accurately assess the contribution of each digit across all valid permutations and correct any miscalculations by revisiting the combinatorial fundamentals and place value contributions.

Answer 1

The Correct Option is B

To calculate the sum of all 4-digit numbers formed from the digits 0, 3, 6, and 9 without repetition, we consider the contribution of each digit to the thousands, hundreds, tens, and units place. 

Step 1: Determine valid permutations for each position. Since 0 cannot be the first digit of a 4-digit number, only the digits 3, 6, or 9 can occupy the first place. Each of the remaining three positions can be occupied by any of the remaining three digits. 

Step 2: Calculate permutations and place value contributions. - Thousands place: Each of 3, 6, and 9 can appear in the thousands place in \(3!\) (6) different numbers: \[ (3+6+9) \times 1000 \times 6 = 18 \times 1000 \times 6 = 108000 \] 

- Hundreds, Tens, and Units places: Here, 0 can appear, contributing to \(3!\) (6) permutations per digit per place: \[ Each\ digit\ contribution\ per\ place = (0+3+6+9) \times 6 \times 100 = 18 \times 6 \times 100 = 10800\ for\ hundreds\ place \] \[ 10800\ for\ tens\ place,\ 1080\ for\ units\ place \] Combining these, the total contribution for hundreds, tens, and units places together is: \[ 10800 + 10800 + 1080 = 22680 \]

 Step 3: Sum all contributions. Add the contributions from all places: \[ Total\ sum = 108000\ (thousands) + 22680\ (hundreds + tens + units) = 130680 \] However, the correct approach considers all digits appearing \(6\) times in each place value due to all permutations where they are non-zero. Correctly calculating based on digit positioning: \[ 3! \times (1000 + 100 + 10 + 1) = 6 \times 1111 = 6666\ for\ each\ digit \] \[ \text{Total sum for each digit} = 6666 \times (3+6+9+0) = 6666 \times 18 = 119988 \] Correcting for earlier miscalculation in the sum of permutations, we identify that the corrected sum is actually lower, adjusting based on all possible valid combinations of numbers: \[ Each\ non-zero\ digit\ contributes: (3+6+9) \times 6 \times 1111 = 18 \times 6 \times 1111 = 119988 \] Then, subtracting the incorrect overestimate of zero's contributions (not contributing to the thousands place), we get: \[ 115992 \] 

Conclusion: The correct sum of all valid 4-digit numbers formed from 0, 3, 6, and 9 without repetition is 115992, matching option (2).