<p>The number of three-digit numbers of the form \(xyz\) such that \(x < y\) and \(z \leq y\) is</p>
Step-by-Step Solution
Key Concept: Recognize that for three-digit number xyz with x < y < z, we need to choose 3 distinct digits from {1,2,...,9} for x, then from {2,3,...,9} for y, and {3,4,...,9} for z while maintaining strict inequality. The key insight is that selecting any 3 distinct digits automatically determines a unique arrangement in increasing order.
<p><strong>Step 1:</strong> Understand the constraint. We need a three-digit number xyz where x < y < z, meaning all three digits are strictly increasing.</p><p><strong>Step 2:</strong> Recognize that x ≥ 1 (since it's the first digit of a three-digit number), and y,z can be any digit from 0-9, but x < y < z must hold.</p><p><strong>Step 3:</strong> Key insight: Every selection of 3 distinct digits from {1,2,3,4,5,6,7,8,9} can be arranged in exactly ONE way to satisfy x < y < z (in increasing order). Since x must be at least 1, we cannot use 0 as the smallest digit.</p><p><strong>Step 4:</strong> Therefore, the problem reduces to: "In how many ways can we choose 3 distinct digits from the set {1,2,3,4,5,6,7,8,9}?"</p><p><strong>Step 5:</strong> This is simply C(9,3) = 9!/(3!×6!) = (9×8×7)/(3×2×1) = 504/6 = 84</p><p>∴ Answer: <strong>B (or the option equal to 84)</strong></p>
Correct Answer: B