<p>How many different seven-digit numbers can be made of distinct digits if the sum of the digits is even?</p>
Step-by-Step Solution
Key Concept: A sum is even iff we use an even count of odd digits. Since we need 7 distinct digits from {0,1,2,...,9} with 5 odd and 5 even digits available, we must use either 1, 3, 5, or 7 odd digits. However, using 7 odd digits is impossible (only 5 exist), so we use 1, 3, or 5 odd digits paired with 6, 4, or 2 even digits respectively. The first digit cannot be 0, requiring careful case analysis.
<p><strong>Step 1: Determine parity condition.</strong> Sum is even ⟺ even count of odd digits among our 7 chosen digits. With 5 odd {1,3,5,7,9} and 5 even {0,2,4,6,8}, we need k ∈ {1,3,5} odd digits.</p><p><strong>Step 2: Count selections for each case.</strong></p><p>• k=1 (1 odd, 6 even): C(5,1)×C(5,6) = 5×0 = 0 [impossible, only 5 evens]</p><p>• k=3 (3 odd, 4 even): C(5,3)×C(5,4) = 10×5 = 50 selections</p><p>• k=5 (5 odd, 2 even): C(5,5)×C(5,2) = 1×10 = 10 selections</p><p><strong>Step 3: Count valid arrangements for k=3 (50 selections).</strong> Total arrangements: 50×7! = 50×5040 = 252,000. Subtract cases with 0 in position 1: If 0 is included in the 4 evens selected, we have C(5,3)×C(4,3) = 10×4 = 40 selections. Each gives 6!×40 = 720×40 = 28,800 arrangements with 0 first. Valid: 252,000 − 28,800 = 223,200.</p><p><strong>Step 4: Count valid arrangements for k=5 (10 selections).</strong> Total: 10×7! = 50,400. None include 0 (all 5 odds, only 2 of 5 evens). Valid: 50,400.</p><p><strong>∴ Answer: 223,200 + 50,400 = 273,600</strong></p>
Correct Answer: 223