<p>The sum of all the numbers of four different digits that can be made by using the digits 0, 1, 2, and 3 is</p>
Step-by-Step Solution
Key Concept: Each digit appears equally often in each position (units, tens, hundreds, thousands) across all valid permutations. Use symmetry: calculate how many times each digit occupies each place value, then sum.
<p><strong>Step 1:</strong> Total 4-digit numbers using 0,1,2,3 (all different): We need arrangements where the first digit ≠ 0.</p><p>Total permutations of 0,1,2,3 = 4! = 24. Invalid (starting with 0) = 3! = 6. Valid 4-digit numbers = 24 - 6 = 18.</p><p><strong>Step 2:</strong> Find frequency of each digit in each position.</p><p><strong>Thousands place:</strong> Only 1,2,3 can be here. Each appears in 3! = 6 numbers. (0 appears 0 times)</p><p><strong>Hundreds/Tens/Units places:</strong> For each of these 3 positions, when we fix one position for digit d, we arrange remaining 3 digits in remaining 3 positions = 3! = 6 ways. But we must exclude cases starting with 0. In each non-thousands place, each digit (0,1,2,3) appears 6 times considering all 24 perms, minus cases where 0 is first. Each digit appears (6 - 2) = 4 times in hundreds place, and similarly 4 times in tens and units places.</p><p><strong>Step 3:</strong> Sum = (Thousands place) + (Hundreds place) + (Tens place) + (Units place)</p><p>= 1000(6×1 + 6×2 + 6×3) + 100(4×0 + 4×1 + 4×2 + 4×3) + 10(4×0 + 4×1 + 4×2 + 4×3) + 1(4×0 + 4×1 + 4×2 + 4×3)</p><p>= 1000(6×6) + 100(4×6) + 10(4×6) + 1(4×6)</p><p>= 1000(36) + 100(24) + 10(24) + 24</p><p>= 36000 + 2400 + 240 + 24</p><p><strong>∴ Answer: 38664 (Option C)</strong></p>
Correct Answer: C