Understanding the Problem
To find the number of arrangements of the letters in the word "FAILURE" where the vowels are always together, we first identify the vowels and consonants in the word.
Vowels and Consonants
- Vowels: A, I, U, E (4 vowels)
- Consonants: F, L, R (3 consonants)
Treating Vowels as a Single Unit
When vowels are kept together, we can treat them as a single unit or block. Thus, we have:
- Vowel block: (AIUE)
- Consonants: F, L, R
This gives us a total of 4 units to arrange: (AIUE), F, L, R.
Arranging the Units
The number of ways to arrange these 4 units (the vowel block and the consonants) is calculated as follows:
- Arrangements of 4 units: 4! = 24 ways
Arranging the Vowels within the Block
Next, we need to arrange the vowels within their block. The vowels A, I, U, E can be arranged in:
- Arrangements of 4 vowels: 4! = 24 ways
Calculating the Total Arrangements
To find the total number of arrangements where vowels are always together, we multiply the arrangements of the units by the arrangements of the vowels:
- Total arrangements = 4! * 4! = 24 * 24 = 576
Final Answer
Thus, the total number of arrangements of the letters in "FAILURE" such that the vowels are always together is 576.