<p>Find the number of odd numbers greater than two million that can be made with the digits of the number 1315414.</p>
Step-by-Step Solution
Key Concept: An odd number must end in an odd digit (1, 3, or 5), and we must form 7-digit numbers greater than 2,000,000. Count the permutations by fixing the last digit as odd, then arrange remaining digits ensuring the first digit is ≥ 2.
<p><strong>Given digits:</strong> 1, 3, 1, 5, 4, 1, 4 → Three 1's, two 4's, one 3, one 5</p><p><strong>Step 1:</strong> For the number to be odd, last digit ∈ {1, 3, 5}</p><p><strong>Step 2:</strong> For the number to be > 2,000,000, first digit must be ≥ 2, so first digit ∈ {3, 4, 5}</p><p><strong>Step 3: Case 1 - Last digit = 1</strong><br/>Remaining digits: 1, 1, 3, 5, 4, 4<br/>First digit must be from {3, 4, 5} (all available)<br/>• If first = 3: Arrange {1, 1, 5, 4, 4} in 5 positions = 5!/(2!·2!) = 30<br/>• If first = 4: Arrange {1, 1, 3, 5, 4} in 5 positions = 5!/2! = 60<br/>• If first = 5: Arrange {1, 1, 3, 4, 4} in 5 positions = 5!/(2!·2!) = 30<br/>Subtotal: 120</p><p><strong>Step 4: Case 2 - Last digit = 3</strong><br/>Remaining digits: 1, 1, 1, 5, 4, 4<br/>First digit must be from {4, 5}<br/>• If first = 4: Arrange {1, 1, 1, 5, 4} in 5 positions = 5!/3! = 20<br/>• If first = 5: Arrange {1, 1, 1, 4, 4} in 5 positions = 5!/(3!·2!) = 10<br/>Subtotal: 30</p><p><strong>Step 5: Case 3 - Last digit = 5</strong><br/>Remaining digits: 1, 1, 1, 3, 4, 4<br/>First digit must be from {3, 4}<br/>• If first = 3: Arrange {1, 1, 1, 4, 4} in 5 positions = 5!/(3!·2!) = 10<br/>• If first = 4: Arrange {1, 1, 1, 3, 4} in 5 positions = 5!/3! = 20<br/>Subtotal: 30</p><p><strong>∴ Answer: 120 + 30 + 30 = 180</strong></p>
Correct Answer: 120