In how many different permutations can the letters of the word THERAPY be arranged such that the two vowels E and A never appear together as adjacent letters?

Introduction / Context:
This problem is about permutations with a restriction that certain letters must not be adjacent. The word THERAPY has two vowels, and we want to count the number of arrangements in which these vowels do not come together as a single block.

Given Data / Assumptions:

• Word: THERAPY.
• Letters: T, H, E, R, A, P, Y.
• Total letters = 7 and all are distinct.
• Vowels: E and A; consonants: T, H, R, P, Y.
• Vowels must not be adjacent in the arrangement.

Concept / Approach:
A standard strategy is to first count all possible permutations without any restriction, and then subtract the number of permutations where the vowels are together as a single block. The result gives the arrangements where the vowels never come together.

Step-by-Step Solution:
Step 1: Total permutations of 7 distinct letters without restriction = 7!. Step 2: Compute 7! = 5040. Step 3: Now count permutations where E and A are together. Treat the pair (E, A) or (A, E) as a single block X. Step 4: Then the objects to arrange are X and the 5 consonants T, H, R, P, Y. So we have 6 distinct objects. Step 5: Arrangements of these 6 distinct objects = 6! = 720. Step 6: Inside the block X, E and A can be arranged as EA or AE, giving 2 internal arrangements. Step 7: Total arrangements with vowels together = 6! * 2 = 720 * 2 = 1440. Step 8: Arrangements where vowels never come together = total permutations minus permutations with vowels together = 5040 - 1440 = 3600.

Verification / Alternative check:
You can also think of placing the consonants first and then inserting vowels in the gaps, but the subtraction method is straightforward and reliable for this case. The calculation 5040 - 1440 clearly leads to 3600 as the count of valid arrangements.

Why Other Options Are Wrong:
1440 is the count of arrangements where the vowels are together, not where they are separated. 720 is 6!, which is insufficient here. 2250 is not linked to any standard counting step in this approach.

Common Pitfalls:
The main error is to count only the cases where vowels are together and forget to subtract from the total. Some also forget the factor of 2 for the internal arrangements of the vowels within the block.

Final Answer:
The number of permutations of THERAPY in which the vowels never appear together is 3600.