To solve this problem, we can use the concept of permutations. Let's break down the solution into different steps.

Step 1: Count the total number of ways to arrange all the letters of the word EAMCET without any restrictions.

The word EAMCET has 6 letters, so there are 6! (6 factorial) ways to arrange all the letters without any restrictions. The factorial of a number is the product of all positive integers less than or equal to that number.

6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

Therefore, there are 720 ways to arrange all the letters of the word EAMCET without any restrictions.

Step 2: Count the number of ways in which the two vowels (E and A) are together.

Since the two vowels must be together, we can treat them as a single entity. So, we have 5 entities to arrange: (EA), M, C, E, T.

The number of ways to arrange these 5 entities is 5!.

5! = 5 x 4 x 3 x 2 x 1 = 120

However, within the (EA) entity, the vowels E and A can be arranged in 2! ways.

2! = 2 x 1 = 2

Therefore, the number of ways in which the two vowels are together is 120 x 2 = 240.

Step 3: Subtract the number of ways in which the two vowels are together from the total number of ways to arrange all the letters.

Total ways without any restrictions - Ways with vowels together = 720 - 240 = 480

Step 4: Count the number of ways in which the two vowels are not together.

Since there are 480 ways in which the two vowels are not together, we can treat this as the total number of ways in which the letters of the word EAMCET can be arranged such that two vowels are never together.

Step 5: Simplify the number of ways by cancelling any common factors.

The number of ways can be simplified by dividing both the numerator and denominator by their common factor, which is 2.

480/2 = 240

Therefore, the number of ways to arrange the letters of the word EAMCET such that two vowels are never together is 240.

So, the correct answer is option C) 72