Permutations & Combinations
Word Formation
Grade 11
Question:
<p>With 17 consonants and 5 vowels, the number of four-letter words that can be formed having 2 vowels in the middle and one consonant repeated or different consonants at each end is</p>
<p>A. 2890</p>
<p>B. 5440</p>
<p>C. 5780</p>
<p>D. none of these</p>
Step-by-Step Solution
Key Concept: The word structure is fixed as C-V-V-C (consonant-vowel-vowel-consonant). The two end positions must have consonants that can be identical or different, while middle positions have vowels with specific repetition rules.
<p><strong>Step 1:</strong> Identify the structure: Position 1 (consonant) - Position 2 (vowel) - Position 3 (vowel) - Position 4 (consonant)</p><p><strong>Step 2:</strong> <strong>Case 1 - Same consonant at both ends (repeated):</strong><br>• Choose 1 consonant from 17 for positions 1 and 4: 17 ways<br>• Choose vowel for position 2 from 5: 5 ways<br>• Choose vowel for position 3 from 5: 5 ways<br>• Subtotal: 17 × 5 × 5 = 425</p><p><strong>Step 3:</strong> <strong>Case 2 - Different consonants at both ends:</strong><br>• Choose consonant for position 1 from 17: 17 ways<br>• Choose consonant for position 4 from remaining 16: 16 ways<br>• Choose vowel for position 2 from 5: 5 ways<br>• Choose vowel for position 3 from 5: 5 ways<br>• Subtotal: 17 × 16 × 5 × 5 = 6800</p><p><strong>Step 4:</strong> Total = Case 1 + Case 2 = 425 + 6800 = <strong>7225</strong></p><p>∴ Answer: D</p>
Correct Answer: D