<p>Among 10 persons, \(A\), \(B\), \(C\) are to speak at a function. The number of ways in which it can be done if \(A\) wants to speak before \(B\) and \(B\) wants to speak before \(C\) is</p>
Step-by-Step Solution
Key Concept: When A must speak before B and B must speak before C (a fixed relative order), we select 3 positions from 10 and arrange them in only ONE way. Use combinations, not permutations, since the relative order is already determined.
<p><strong>Step 1:</strong> We need to place A, B, C in 3 of the 10 positions such that A comes before B, and B comes before C.</p><p><strong>Step 2:</strong> Choose any 3 positions from 10 positions: C(10,3) ways.</p><p><strong>Step 3:</strong> Once 3 positions are selected, there is exactly ONE way to arrange A, B, C in these positions while maintaining A before B before C (A goes in the earliest position, B in the middle, C in the latest).</p><p><strong>Step 4:</strong> Calculate: C(10,3) = 10!/(3!×7!) = (10×9×8)/(3×2×1) = 720/6 = 120</p><p>∴ Answer: <strong>C (120)</strong></p>
Correct Answer: C