<p>How many words can be made with the letters of the word INSTITUTION such that vowels and consonants alternate?</p>
Step-by-Step Solution
Key Concept: Identify vowels (I, U, I, O) and consonants (N, S, T, T, T, N) separately, then recognize that with 4 vowels and 6 consonants, alternation is impossible in a standard arrangement—unless the pattern specifically accounts for which letter type must start and end.
<p><strong>Step 1:</strong> Identify letters in INSTITUTION</p><p>Vowels: I, I, U, O (4 vowels, with I repeated twice)</p><p>Consonants: N, S, T, T, T, N (6 consonants: N appears 2 times, T appears 3 times)</p><p><strong>Step 2:</strong> Check alternation feasibility</p><p>For perfect alternation in a 10-letter word: need pattern CVCVCVCVCVC (6 consonants, 5 vowels) OR VCVCVCVCVCV (5 consonants, 5 vowels)</p><p>Since we have 6 consonants and 4 vowels, only CVCVCVCVCVC works (using 6 consonants, 4 vowels).</p><p><strong>Step 3:</strong> Arrange consonants in 6 positions</p><p>Consonants N, N, S, T, T, T in 6 positions = 6!/(2!×3!) = 720/12 = 60</p><p><strong>Step 4:</strong> Arrange vowels in 5 remaining positions</p><p>Choose 4 of 5 positions for vowels I, I, U, O = C(5,4) × 4!/(2!) = 5 × 24/2 = 60</p><p><strong>Step 5:</strong> Total arrangements</p><p>60 × 60 = 3600</p><p>∴ Answer: <strong>3600</strong></p>
Correct Answer: 3600