In how many different ways can the letters of the word OPTICAL be arranged so that all the vowels in the word always appear together in every arrangement?

Introduction / Context:
This is another grouping problem involving permutations with a condition on vowels. The word OPTICAL contains both vowels and consonants, and we are asked to count arrangements where all the vowels appear side by side. Such questions test your ability to restructure the problem by temporarily treating a group of letters as a single combined letter and then using factorials to count arrangements.

Given Data / Assumptions:
The word is OPTICAL. Letters are O, P, T, I, C, A, L. Vowels: O, I, A. Consonants: P, T, C, L. All seven letters are distinct. Vowels must always appear together as one contiguous block.

Concept / Approach:
We group the vowels O, I and A into a single super letter. Then we have this block plus the consonants P, T, C and L to arrange. So, we first count how many ways to arrange these 5 distinct objects. Next, we count how many ways the vowels can arrange among themselves inside the block. Using the multiplication principle, we combine these counts to reach the final number of valid arrangements.

Step-by-Step Solution:
Step 1: Treat O, I and A as one block V. Step 2: The objects to arrange are V, P, T, C and L. Thus we have 5 distinct objects. Step 3: The number of ways to arrange 5 distinct objects in a row is 5! = 120. Step 4: Inside block V, the vowels O, I and A can be arranged in 3! = 6 different ways. Step 5: For each arrangement of the 5 objects, there are 6 possible internal arrangements of the vowels. Step 6: Total number of required arrangements = 5! * 3! = 120 * 6 = 720.

Verification / Alternative check:
Without any restriction, all the letters of OPTICAL can be arranged in 7! = 5040 ways. The condition that vowels must appear together clearly restricts the arrangements, so the answer should be significantly smaller than 5040. We found 720 valid arrangements, which is exactly 5040 divided by 7. This is reasonable because the vowel block behaves like a single letter among seven possible positions, and the internal arrangements of vowels adjust the count appropriately. A quick sanity check of the arithmetic also confirms that 5! = 120 and 3! = 6 give 720 when multiplied.

Why Other Options Are Wrong:
The value 360 would arise if someone incorrectly used 4! instead of 5! or made a similar partial counting error. Values like 700 or 120 are not of the factorial form needed here, and 120 corresponds to only the outer arrangements of the vowel block and consonants, ignoring the internal arrangements of the vowels. Therefore they all underestimate the total number of valid permutations. Only 720 correctly combines both outer and inner arrangements.

Common Pitfalls:
Frequently, students forget to consider the internal permutations of the grouped vowels, which leads them to answer 5! instead of 5! * 3!. Another issue is mistakenly splitting the vowel block in the middle during counting, which violates the requirement that vowels stay together. Also, some learners incorrectly treat vowels as indistinguishable, even though O, I and A are different letters. Carefully separating outer block arrangements from inner vowel arrangements prevents these mistakes.

Final Answer:
The number of different arrangements of OPTICAL in which all vowels always appear together is 720.