<p>In an examination of nine papers, a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is</p>
Step-by-Step Solution
Key Concept: A candidate is unsuccessful when failures ≥ passes. Since there are 9 papers total, if f = failures and p = passes, then f + p = 9 and unsuccessful means f ≥ p, which gives f ≥ 4.5, so f ≥ 5. Count all such cases.
<p><strong>Step 1:</strong> Define the condition for being unsuccessful. A candidate is unsuccessful when: passes ≤ failures, or equivalently, failures ≥ passes.</p><p><strong>Step 2:</strong> With 9 papers total: failures + passes = 9. For unsuccessful: f ≥ p, so f ≥ 9 - f, giving 2f ≥ 9, thus f ≥ 4.5, which means f ≥ 5.</p><p><strong>Step 3:</strong> Count ways for unsuccessful cases (f = 5, 6, 7, 8, 9):</p><p>• f = 5, p = 4: C(9,5) = 126</p><p>• f = 6, p = 3: C(9,6) = 84</p><p>• f = 7, p = 2: C(9,7) = 36</p><p>• f = 8, p = 1: C(9,8) = 9</p><p>• f = 9, p = 0: C(9,9) = 1</p><p><strong>Step 4:</strong> Total unsuccessful ways = 126 + 84 + 36 + 9 + 1 = 256</p><p>Alternatively: Total outcomes = 2⁹ = 512, and by symmetry, unsuccessful = 512/2 = 256 (excluding the impossible f = p case since 9 is odd)</p><p>∴ Answer: 256</p>
Correct Answer: A