In how many distinct ways can all the letters of the word EDUCATION be rearranged to form new words, if every vowel must always occupy a vowel position and every consonant must always occupy a consonant position exactly as in the original word EDUCATION?

Introduction / Context:
This question tests the concept of permutations with positional restrictions. The word EDUCATION contains both vowels and consonants, and we want to rearrange its letters while keeping the pattern of vowel positions and consonant positions the same as in the original word. Such problems are common in aptitude tests to check understanding of permutations and classification of letters into groups like vowels and consonants.

Given Data / Assumptions:

• The word is EDUCATION.
• The letters are E, D, U, C, A, T, I, O, N.
• Vowels in this word are E, U, A, I, O.
• Consonants in this word are D, C, T, N.
• The relative pattern of vowel and consonant positions must remain exactly the same as in EDUCATION.
• All letters are used in every arrangement.

Concept / Approach:
The key idea is that vowels can only occupy positions which were originally vowels in EDUCATION, and consonants can only occupy positions which were originally consonants. Once we fix which positions are for vowels and which are for consonants, we simply count how many ways to permute the vowels among vowel positions and the consonants among consonant positions. Since the choice of vowels is independent of the choice of consonants, the total number of arrangements is the product of the two permutation counts.

Step-by-Step Solution:
Step 1: Identify positions of vowels in EDUCATION. The word is E D U C A T I O N, so vowels appear at positions 1 (E), 3 (U), 5 (A), 7 (I) and 8 (O). That gives 5 vowel positions.Step 2: Identify positions of consonants. Consonants D, C, T, N appear at positions 2, 4, 6 and 9. That gives 4 consonant positions.Step 3: Count permutations of the 5 distinct vowels among the 5 vowel positions. This is 5! ways.Step 4: Count permutations of the 4 distinct consonants among the 4 consonant positions. This is 4! ways.Step 5: Multiply the independent counts for vowels and consonants. Total arrangements = 5! * 4!.Step 6: Compare with the given options. The expression 5! * 4! matches option written as 4! x 5!.

Verification / Alternative check:
An alternative way to think about the problem is to imagine two separate slots: one row of vowel slots and one row of consonant slots. First, arrange the 5 vowels in all possible orders, which gives 5! permutations. Then separately arrange the 4 consonants, which gives 4! permutations. Because every full word must be obtained by combining one vowel arrangement with one consonant arrangement, again the total count is 5! * 4!. There is no other restriction, so this confirms the result.

Why Other Options Are Wrong:

• 4! x 4!: This would imply that there are only 4 vowels and 4 consonants, which is incorrect because there are 5 vowels in EDUCATION.
• 5! x 5!: This assumes 5 vowels and 5 consonants, but the word has only 4 consonants, so this overcounts arrangements.
• 3! x 4!: This assumes only 3 vowel positions being permuted, which does not match the actual structure of the word.

Common Pitfalls:
Students often forget to classify letters correctly and may miscount vowels or consonants. Another common mistake is to think that the pattern of actual letters must remain fixed, instead of the pattern of vowel and consonant positions. Yet another error is to simply take 9! permutations without considering the restriction on positions. Some learners also confuse 5! + 4! with 5! * 4!, but in problems where two independent choices are made, multiplication is the correct rule of counting.

Final Answer:
The total number of valid rearrangements is 4! x 5!, which corresponds to 5! vowel permutations times 4! consonant permutations.