Permutations & Combinations
Counting problems
Grade 11

Question:

<p>In an examination a candidate has to pass in each of the papers. If the total number of different ways in which the candidate can fail is 63 then find the number of papers in the examination.</p>

Step-by-Step Solution

Key Concept: If there are n papers and a candidate must fail in at least one paper to 'fail overall', then total failing ways = (2^n - 1), where 2^n represents all possible pass/fail combinations for n papers and we subtract 1 to exclude the 'pass all' case.
<p><strong>Step 1:</strong> For each paper, the candidate can either pass or fail (2 choices per paper).</p><p><strong>Step 2:</strong> For n papers, total possible outcomes = 2^n</p><p><strong>Step 3:</strong> This includes the case where the candidate passes all papers (only 1 way). Since the candidate must fail at least one paper, we exclude this case.</p><p><strong>Step 4:</strong> Number of ways to fail = 2^n - 1</p><p><strong>Step 5:</strong> Given: 2^n - 1 = 63</p><p><strong>Step 6:</strong> 2^n = 64 = 2^6</p><p><strong>Step 7:</strong> n = 6</p><p>∴ Answer: 6</p>
Correct Answer: 6

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