Permutations & Combinations
Words with repetition
Grade None
Question:
<p>Words of length 10 are formed using the letters \(A, B, C, D, E, F, G, H, I, J\). Let \(x\) be the number of such words where no letter is repeated; and let \(y\) be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then \(\dfrac{y}{9x}\) = ________.</p>
Step-by-Step Solution
Key Concept: Calculate x as permutations of 10 distinct letters (10!), then y by choosing which letter repeats (10 ways), choosing 8 other letters from remaining 9, and arranging all 11 positions with one repeated letter (10·C(9,8)·11!/2!). The ratio y/(9x) simplifies to a clean integer.
<p><strong>Step 1: Find x (no letter repeated)</strong></p><p>All 10 distinct letters A, B, C, D, E, F, G, H, I, J must be used exactly once in a 10-letter word.</p><p>x = 10!</p><p><strong>Step 2: Find y (exactly one letter repeated twice)</strong></p><p>We need a 10-letter word where one letter appears exactly twice and the other 8 letters are distinct and appear once each.</p><p>• Choose which letter repeats: 10 ways</p><p>• Choose 8 letters from remaining 9 letters: C(9,8) = 9 ways</p><p>• Arrange these 10 letters (with one letter appearing twice): 10!/2! positions</p><p>y = 10 × 9 × (10!/2!) = 90 × (10!/2)</p><p><strong>Step 3: Calculate the ratio</strong></p><p>$$ rac{y}{9x} = rac{90 × rac{10!}{2}}{9 × 10!} = rac{90}{9 × 2} = rac{90}{18} = 5$$</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: 5