<p>Total number of six-digit numbers that can be formed having the property that every succeeding digit is greater than the preceding digit is equal to</p>
Step-by-Step Solution
Key Concept: A six-digit number where each digit is strictly greater than the previous one is equivalent to selecting 6 distinct digits from {1,2,3,4,5,6,7,8,9} and arranging them in increasing order. Since increasing order is unique, this becomes a pure combination problem: C(9,6).
<p><strong>Step 1:</strong> Identify the constraint. We need six-digit numbers where d₁ < d₂ < d₃ < d₄ < d₅ < d₆, with d₁ ≥ 1 (valid six-digit number).</p><p><strong>Step 2:</strong> Recognize that any selection of 6 distinct digits from {1,2,3,4,5,6,7,8,9} automatically arranges in exactly one way (increasing order). We exclude 0 because if 0 is selected, it must come first, violating d₁ ≥ 1.</p><p><strong>Step 3:</strong> The problem reduces to: select 6 distinct digits from the 9 available digits {1,2,3,4,5,6,7,8,9}.</p><p><strong>Step 4:</strong> Number of ways = C(9,6) = 9!/(6!×3!) = (9×8×7)/(3×2×1) = 504/6 = 84</p><p>∴ Answer: <strong>84</strong> (Option B)</p>
Correct Answer: B