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