**Arranging the Letters in the Word "FAILURE"**

To find the number of arrangements of the letters in the word "FAILURE" where the vowels are always together, we can treat the vowels ("A", "I", and "U") as a single entity. This means that we have to arrange the consonants ("F", "L", "R", and "E") along with this group of vowels.

Let's consider the group of vowels as a single letter "V". So, now we have the word "F V L R E".

**Arranging the Consonants**

The consonants "F", "L", "R", and "E" can be arranged in 4! = 24 ways.

**Arranging the Group of Vowels**

Now, the group of vowels "V" (which represents "A", "I", and "U") can be arranged among themselves in 3! = 6 ways.

**Arranging the Consonants and the Group of Vowels**

Since the group of vowels is treated as a single letter, we can consider the arrangement of "F V L R E" as arranging 5 letters where "V" is repeated.

The total number of arrangements of the consonants and the group of vowels is 5! = 120.

**Considering the Repetition of "V"**

In the above calculation, we have considered the "V" as a single letter, but it actually represents three vowels. So, we need to consider the repetition of "V" in the arrangements.

The number of ways to arrange the vowels among themselves is 3! = 6, but within each of these arrangements, the vowels can be rearranged among themselves in 3! = 6 ways. So, we need to multiply the number of arrangements by 6.

**Calculating the Final Answer**

To find the number of arrangements where the vowels are always together, we multiply the number of arrangements of the consonants and the group of vowels by the number of ways to arrange the vowels among themselves:

Number of arrangements = 120 * 6 = 720.

Therefore, there are 720 different arrangements of the letters in the word "FAILURE" where the vowels are always together.

4!×⁴p3 = 24×23 =576