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, A, E, I (4 vowels, with 2 identical A's).
• Consonants: M, M, T, T, H, C, S (7 consonants, with 2 identical M's and 2 identical T's).

Step 2: Arrange the 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 the first consonant, between each pair, after the last consonant). We need to choose 4 gaps to place vowels:
[ \binom{8}{4} = 70 ]

Step 4: Arrange the Vowels

The number of distinct arrangements of vowels (A,A,E,I) is:
[ \frac{4!}{2!} = \frac{24}{2} = 12 ]

Step 5: Total Arrangements

Multiply all results:
[ 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。