<p>Find the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time.</p>
Step-by-Step Solution
Key Concept: Each digit appears in each position (units, tens, hundreds, thousands) equally often across all permutations. Use symmetry: each digit occupies each place exactly (4-1)! = 6 times, then sum = (sum of digits) × (frequency per position) × (place value sum).
<p><strong>Step 1:</strong> Total permutations of digits 2, 3, 4, 5 taken all at a time = 4! = 24</p><p><strong>Step 2:</strong> By symmetry, each digit appears in each position (units, tens, hundreds, thousands) equally often.</p><p>Frequency of each digit in each position = 24 ÷ 4 = 6 times</p><p><strong>Step 3:</strong> Sum of the digits = 2 + 3 + 4 + 5 = 14</p><p><strong>Step 4:</strong> Contribution from each position:</p><p>• Units place: 14 × 6 × 1 = 84</p><p>• Tens place: 14 × 6 × 10 = 840</p><p>• Hundreds place: 14 × 6 × 100 = 8400</p><p>• Thousands place: 14 × 6 × 1000 = 84000</p><p><strong>Step 5:</strong> Total sum = 84 + 840 + 8400 + 84000 = <strong>93324</strong></p><p><strong>Alternatively:</strong> Sum = (sum of digits) × (frequency per position) × (1 + 10 + 100 + 1000) = 14 × 6 × 1111 = 93324</p>
Correct Answer: 93324