Number of arrangements of the word CAPITAL so that all vowels remain together in every arrangement

Introduction / Context:
This problem focuses on permutations of letters in a word with the additional condition that all vowels must appear together as one block. It combines the idea of treating a group of letters as a single unit and then arranging letters inside that unit.

Given Data / Assumptions:

• Word: CAPITAL.
• Letters: C, A, P, I, T, A, L.
• Vowels: A, I, A.
• Consonants: C, P, T, L.
• All vowels must appear together in every arrangement.
• Letters A appear twice and are identical.

Concept / Approach:
First we treat the group of vowels as a single block so that they always remain together. Then we arrange this block together with the consonants. Inside the vowel block, we also need to count the permutations of the vowels, taking into account the repeated letter A.

Step-by-Step Solution:
Step 1: Group the three vowels A, A, I as a single block, say V.Step 2: Now we have the block V and the four consonants C, P, T, L.Step 3: So we have 5 items in total: V, C, P, T, L.Step 4: These 5 distinct items can be arranged in 5! ways, so we have 5! = 120 arrangements of the block and consonants.Step 5: Inside the vowel block V, the letters are A, A, I.Step 6: If they were all distinct, we would have 3! permutations, but the letter A is repeated twice.Step 7: Number of permutations inside the block = 3! / 2! = 6 / 2 = 3.Step 8: Total valid arrangements = arrangements of blocks * internal vowel arrangements = 120 * 3 = 360.

Verification / Alternative check:
You can cross check by listing the three internal vowel arrangements, such as A A I, A I A, and I A A, and then imagine placing this block in 5! ways among the consonants. Since each internal arrangement can be paired with 120 external arrangements, the total must be 360, which confirms our calculation.

Why Other Options Are Wrong:

• 720 would correspond to 6! for some miscounted set, which ignores repetition of A and the block idea.
• 120 counts only the arrangements of the block and consonants without considering internal vowel permutations.
• 840 does not match any standard factorial based combination of these counts.

Common Pitfalls:
Students often forget to divide by 2! for the repeated A, leading to 5! * 3! which equals 720. Another error is to arrange all 7 letters first and then try to subtract cases where vowels are not together, which becomes complicated. Treating the vowels as one block from the start is the clean and simple method.

Final Answer:
The number of valid arrangements of CAPITAL with all vowels together is 360, so the correct answer is 360.