<p>Consider the letters of the word MATHEMATICS. Possible number of words in which no two vowels are together is</p>
<p>\(7! \cdot {}^8C_4 \cdot \dfrac{4!}{2!}\)</p>
<p>\(\dfrac{7!}{2!} \cdot {}^8C_4 \cdot \dfrac{4!}{2!}\)</p>
<p>\(\dfrac{7!}{2!2!} \cdot {}^8C_4 \cdot \dfrac{4!}{2!}\)</p>
<p>\(\dfrac{7!}{2!2!2!} \cdot {}^8C_4 \cdot \dfrac{4!}{2!}\)</p>
Step-by-Step Solution
Key Concept: Arrange consonants first to create gaps, then place vowels in those gaps to ensure separation. The number of gaps available for vowels equals one more than the number of consonants.
<p><strong>Step 1: Identify vowels and consonants in MATHEMATICS</strong></p><p>Vowels: A, A, E, I (4 vowels: 2 identical A's)</p><p>Consonants: M, T, H, M, T, C, S (7 consonants: M appears 2 times, T appears 2 times)</p><p><strong>Step 2: Arrange consonants first</strong></p><p>Number of ways to arrange 7 consonants with M repeated twice and T repeated twice:</p><p>= 7!/(2! × 2!) = 5040/4 = 1260</p><p><strong>Step 3: Identify gaps for vowel placement</strong></p><p>When 7 consonants are arranged in a line, they create 8 gaps (before first, between each pair, after last):</p><p>_C_C_C_C_C_C_C_</p><p>To ensure no two vowels are together, we must place vowels in different gaps.</p><p><strong>Step 4: Select gaps and arrange vowels</strong></p><p>Choose 4 gaps from 8 available gaps: C(8,4) = 70</p><p><strong>Step 5: Arrange the 4 vowels in selected gaps</strong></p><p>Number of ways to arrange 4 vowels (A, A, E, I) where A is repeated twice:</p><p>= 4!/2! = 24/2 = 12</p><p><strong>Step 6: Apply multiplication principle</strong></p><p>Total arrangements = 1260 × 70 × 12</p><p>= 1260 × 840 = 1,058,400</p><p>∴ Answer: <strong>1,058,400</strong> (or equivalent: 1260 × 70 × 12)</p>
Correct Answer: C