Permutations & Combinations
Inclusion-Exclusion principle
Grade 11
Question:
<p>An ordinary cubical dice having six faces marked with alphabets <em>A</em>, <em>B</em>, <em>C</em>, <em>D</em>, <em>E</em>, and <em>F</em> is thrown <em>n</em> times and the list of <em>n</em> alphabets showing up are noted. Find the total number of ways in which among the alphabets <em>A</em>, <em>B</em>, <em>C</em>, <em>D</em>, <em>E</em>, and <em>F</em> only three of them appear in the list.</p>
Step-by-Step Solution
Key Concept: Use inclusion-exclusion principle: first choose which 3 letters appear (6C3 ways), then count n-length sequences using exactly those 3 letters by subtracting cases where fewer than 3 appear.
<p><strong>Step 1:</strong> Choose which 3 alphabets (out of A, B, C, D, E, F) will appear in the list: <strong>^6C_3 ways</strong></p><p><strong>Step 2:</strong> For a fixed set of 3 chosen letters, count sequences of length n where all 3 letters appear at least once.</p><p><strong>Step 3:</strong> Using inclusion-exclusion on the 3 chosen letters:</p><ul><li>Total sequences using only these 3 letters: <strong>3^n</strong></li><li>Subtract sequences missing at least one letter: <strong>^3C_1 · 2^n</strong></li><li>Add back sequences missing at least two letters: <strong>^3C_2 · 1^n = ^3C_2</strong></li><li>Subtract sequences missing all three: <strong>^3C_3 · 0^n = 0</strong></li></ul><p><strong>Step 4:</strong> By inclusion-exclusion, sequences using exactly all 3 letters:</p><p>3^n − ^3C_1(2^n) + ^3C_2(1) = 3^n − 3(2^n) + 3</p><p>Simplify: 3^n − 3·2^n + 3 = 3^n − 3(2^n − 1)</p><p><strong>Step 5:</strong> Multiply by the number of ways to choose 3 letters:</p><p>∴ Answer: <strong>^6C_3 × [3^n − ^3C_2(2^n − 2) − 3]</strong></p><p><em>Note: The form ^3C_2(2^n − 2) represents 3(2^n − 2), which equals 3·2^n − 6, accounting for the proper inclusion-exclusion balance.</em></p>
Correct Answer: \(^{6}C_{3}\times[3^{n} - ^{3}C_{2}(2^{n}-2)-3]\)