How many different arrangements can be made out of the letters of the word DRAUGHT if the two vowels in the word are never separated, that is, the vowels always occur together as a single block in every arrangement?

Introduction / Context:
This is a permutations question involving a restriction on how vowels can appear. We must arrange the letters of a word under the condition that the vowels are always together. Such problems are common in competitive exams and are a direct application of treating a group of letters as a single block to simplify the arrangement count.

Given Data / Assumptions:

• The word is DRAUGHT, containing the letters D, R, A, U, G, H and T.
• Total letters: 7, all distinct.
• The vowels in the word are A and U.
• The condition is that the vowels must never be separated; they must appear together in every arrangement.

Concept / Approach:
When vowels must stay together, we treat the group of vowels as one combined block. This reduces the problem to arranging that block along with the remaining consonants. After arranging the blocks, we then count the internal arrangements of the letters within the vowel block. The total number of valid arrangements is the product of these two counts.

Step-by-Step Solution:
Step 1: Identify vowels: A and U are vowels; the remaining letters D, R, G, H and T are consonants.Step 2: Treat the pair of vowels AU as one block. Along with the five consonants D, R, G, H and T, we have 6 objects to arrange: [AU], D, R, G, H, T.Step 3: The number of ways to arrange 6 distinct objects is 6!.Step 4: Compute 6! = 720.Step 5: Inside the vowel block, A and U can be arranged in 2! ways: AU or UA.Step 6: Total arrangements where the vowels stay together = 6! × 2! = 720 × 2 = 1440.
Verification / Alternative check:
If there were no restriction, the total number of arrangements of 7 distinct letters would be 7! = 5040. We have restricted our arrangements to those where the two vowels are together, giving 1440 possibilities. This is a subset of the full 5040 and seems reasonable in size. A manual check on a smaller analogous example supports the logic of treating the vowels as one block.

Why Other Options Are Wrong:
720 would correspond to the case where we treat the vowel block as a single object but do not consider that the vowels could be internally permuted. 360 and 640 arise from partial or incorrect application of factorials. Only 1440 correctly multiplies the arrangements of the blocks by the internal arrangements of the vowels.

Common Pitfalls:
Students often forget to multiply by the number of internal arrangements of the vowels after treating them as a block. Another mistake is misidentifying the vowels or miscounting the total number of letters. Carefully listing letters and following the block method step by step avoids these errors.

Final Answer:
The number of arrangements where the vowels in DRAUGHT are never separated is 1440.