How many ways can the letters of the word EDUCATION be arranged such that the…

Concept: When the positions in an arrangement are split into two fixed groups — say, the slots currently holding vowels and the slots currently holding consonants — and each letter must stay within its own group's slots, the letters of one group can be freely permuted among themselves independently of the other group. Because the two groups' internal arrangements are independent choices, the total number of valid arrangements is the product of each group's internal permutation count, i.e. (size of group 1)! × (size of group 2)!.

Application: The word EDUCATION has 9 letters, all distinct: E, D, U, C, A, T, I, O, N.

1. Identify the vowels and consonants: the vowels are E, U, A, I, O — 5 letters; the consonants are D, C, T, N — 4 letters.

2. Since relative positions must stay the same, every vowel must remain in one of the 5 positions currently occupied by a vowel, and every consonant must remain in one of the 4 positions currently occupied by a consonant.

3. Within the 5 vowel positions, the 5 distinct vowels can be arranged in 5! ways.

4. Within the 4 consonant positions, the 4 distinct consonants can be arranged in 4! ways.

5. Since the choice made within the vowel group does not restrict the choice within the consonant group, the two counts multiply: 5! × 4! = 120 × 24 = 2880.

Cross-check: The two group sizes (5 vowels + 4 consonants) add up to 9, matching the word's total letter count, so no letter has been double-counted or omitted. All 9 letters of EDUCATION are distinct, so no further division for repeated letters is needed. This confirms 2880 as the total number of arrangements.