Solution:
To solve this question, we need to first understand the condition given in the question: "vowels are always coming together." This means that the letters A, E, and I must always appear together in any arrangement. Therefore, we can consider these three letters as a single unit, which we can call "V."

Now, we have the following letters: F, L, R, V, U. We need to arrange these five letters in different ways.

Step 1: Count the total number of arrangements of the five letters without any restrictions. This can be done using the formula for permutations:

nPr = n!/(n-r)!

where n is the total number of objects and r is the number of objects we are selecting.

In this case, n = 5 and r = 5, so we have:

5P5 = 5!/(5-5)! = 5! = 120

Therefore, there are 120 arrangements of the letters F, L, R, V, and U.

Step 2: Count the number of arrangements where the three vowels (A, E, and I) appear together. Since we have considered these three letters as a single unit (V), we need to arrange the four units F, L, R, U, and V. This can be done using the same formula for permutations:

nPr = n!/(n-r)!

where n = 4 and r = 4, so we have:

4P4 = 4!/(4-4)! = 4! = 24

However, the three vowels (A, E, and I) can appear in any order within the unit V. Therefore, we need to multiply the above result by the number of arrangements of the three vowels, which is 3! = 6.

Therefore, the total number of arrangements where the three vowels appear together is:

24 x 6 = 144

Step 3: Count the number of arrangements where the three vowels do not appear together. To do this, we subtract the number of arrangements where the three vowels appear together from the total number of arrangements without any restrictions:

120 - 144 = -24

This result is negative, which means that there are no arrangements where the three vowels do not appear together.

Step 4: Count the number of arrangements where the vowels (A, E, and I) always appear together. Since we have considered these three letters as a single unit (V), we need to arrange the two units F, L, R, and U, and the unit V. This can be done using the same formula for permutations:

nPr = n!/(n-r)!

where n = 3 and r = 3, so we have:

3P3 = 3!/(3-3)! = 3! = 6

Therefore, there are 6 arrangements of the unit V and 2 arrangements of the remaining units.

Total number of arrangements = 6 x 2 = 12

Step 5: Add the number of arrangements where the vowels always appear together to the number of arrangements where the three vowels appear together:

12 + 144 = 156

Therefore, the total number of arrangements of the letters in the word 'FAILURE', so that vowels are always coming together is 156.

However, we have considered the three vowels as a single unit (V) in our calculations

FAILURE
vowels= a,i,u,e in word
con. = f,l,r

vowels comes together = 4!

total no of words = 4!× 4!= 24×24=576