To solve the problem of finding the number of ways to arrange the letters in "MATHEMATICS" such that no two vowels are adjacent, follow these steps:

Step 1: Identify Vowels and Consonants

• Vowels: A, E, A, I (4 vowels; 2 A's, others unique).
• Consonants: M, T, H, M, T, C, S (7 consonants; 2 M's, 2 T's, others unique).

Step 2: Arrange Consonants

The number of distinct arrangements of consonants is:
[ \frac{7!}{2! \cdot 2!} = \frac{5040}{4} = 1260 ]

Step 3: Choose Gaps for Vowels

7 consonants create 8 gaps (before first, between each pair, after last). We need to choose 4 gaps to place vowels:
[ \binom{8}{4} = 70 ]

Step 4: Arrange Vowels

Vowels have 2 duplicates (A), so the number of distinct arrangements is:
[ \frac{4!}{2!} = \frac{24}{2} = 12 ]

Step 5: Total Arrangements

Multiply all values:
[ 1260 \times 70 \times 12 = 1058400 ]

Answer: (\boxed{1058400})

作者声明:本文包含人工智能生成内容。