In how many different ways can the letters of the word 'THERAPY' be arranged so that the two vowels always come together as a single block in every arrangement?

Introduction / Context:
This problem tests permutations of letters of a word with a specific constraint on vowels. The word THERAPY consists of seven distinct letters, and the condition is that the vowels must always come together in every arrangement. Such questions are standard in permutation and combination topics and help you understand how to treat a group of items as a single block while still allowing internal rearrangements inside that block.

Given Data / Assumptions:
- Word: THERAPY.
- Total letters: 7 (all distinct).
- Vowels in the word: E and A (2 vowels).
- Consonants: T, H, R, P, Y (5 consonants).
- Vowels must always appear together, without any consonant in between them.
- Order of letters matters (we are counting different permutations).

Concept / Approach:
The central idea is to treat the block of vowels as a single combined unit initially. If the vowels are forced to be together, we can think of the pair (E, A) as one object, along with the remaining consonants. After arranging this set of objects, we then count how many internal arrangements are possible for the vowels inside their block. The total number of valid arrangements is obtained by multiplying these two counts, because every arrangement of blocks can be paired with every internal arrangement of the vowels.

Step-by-Step Solution:
Step 1: Group the two vowels E and A together as a single unit, say [EA].Step 2: Now the units to arrange are: [EA], T, H, R, P, Y. This gives 6 distinct units in total.Step 3: The number of ways to arrange 6 distinct units in a row is 6!.Step 4: Compute 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.Step 5: Within the vowel block, the letters E and A can themselves be arranged in 2! ways (EA or AE).Step 6: Compute 2! = 2.Step 7: For each arrangement of the 6 units, there are 2 internal arrangements of the vowels, so total arrangements = 6! * 2! = 720 * 2 = 1440.

Verification / Alternative check:
An alternative way is to recognise that once vowels must stay together, you have effectively reduced the seven-letter problem to a six-letter one with an extra factor for internal vowel permutations. Since the letters are all distinct and there are no repeated letters, no further adjustments are needed. Recomputing 6! as 720 and multiplying again by 2 confirms the total as 1440. This agrees with the systematic method described above and fits the requirement that each valid arrangement has the vowels adjacent as one block.

Why Other Options Are Wrong:
- 720 counts only the permutations of six units and ignores the two internal arrangements of the vowels, undercounting by a factor of 2.
- 1800 is larger than 1440 and does not correspond to any correct factorisation of the problem; it would require 6! * 2.5, which has no combinatorial meaning here.
- 3600 is an overestimate and might come from multiplying by an incorrect extra factor, such as assuming more vowels or miscounting positions.

Common Pitfalls:
Students often forget that when a block is formed, its internal permutations must also be counted. Another frequent mistake is to assume the vowels are fixed in a single internal order, effectively taking only 1 arrangement instead of 2!. Additionally, some learners mistakenly treat the problem as unrestricted permutations of all seven letters (7! = 5040), failing to enforce the adjacency condition. Carefully separating the arrangement of blocks from internal arrangements prevents such errors.

Final Answer:
The letters of the word THERAPY can be arranged in 1440 different ways such that the vowels always come together.