How many different arrangements of all the letters of the word COMMITTEE are possible in which the four vowels in the word do not all appear together as a single block?

Introduction / Context:
This problem involves permutations of a word with repeated letters, under a restriction on the placement of vowels. The word COMMITTEE contains repeated letters M, T and E, which means we must use permutations of multiset formulae. We first count all possible arrangements and then subtract the arrangements where all four vowels are together. This is a standard use of the complement method in permutation and combination problems.

Given Data / Assumptions:
Word: COMMITTEE. Letters: C, O, M, M, I, T, T, E, E. Vowels: O, I, E, E (four vowels with E repeated twice). Consonants: C, M, M, T, T. We consider all 9 letters in each arrangement. Condition: the four vowels must not all appear together as one single block.

Concept / Approach:
The total number of distinct permutations of COMMITTEE is found by dividing 9! by factorials of the counts of repeated letters. To enforce the condition that vowels do not all sit together, we compute two values: total arrangements and arrangements where all vowels are grouped as a single block. The required count is total minus the grouped vowel count. For the grouped case, we glue the four vowels into one super block and then treat the multiset of this block with the consonants, taking care of repeated consonants and internal permutations of vowels.

Step-by-Step Solution:
Step 1: Count total permutations of COMMITTEE. Step 2: There are 9 letters with repetitions: M appears twice, T appears twice, E appears twice. Step 3: Total arrangements = 9! / (2! * 2! * 2!). Step 4: Compute 9! = 362880 and 2! * 2! * 2! = 8. So total arrangements = 362880 / 8 = 45360. Step 5: Now count arrangements where all vowels O, I, E, E form one block. Step 6: Treat the vowel block V as one item. Together with consonants C, M, M, T, T we have 6 items: V, C, M, M, T, T. Step 7: Among these 6 items, M is repeated twice and T is repeated twice. Step 8: Arrangements of these 6 items = 6! / (2! * 2!) = 720 / 4 = 180. Step 9: Inside the vowel block V, the letters are O, I, E, E where E is repeated twice. Step 10: Internal arrangements of vowels = 4! / 2! = 24 / 2 = 12. Step 11: Total arrangements with all vowels together = 180 * 12 = 2160. Step 12: Required arrangements with vowels not all together = total - grouped = 45360 - 2160 = 43200.

Verification / Alternative check:
A reasonable check is to compare magnitudes. The number of grouped arrangements (2160) is relatively small compared to the total 45360, which makes sense because enforcing all four vowels to be together is a strict restriction. Subtracting 2160 leaves 43200, which is still large but slightly less than the full total. Also, computing with a calculator or careful manual arithmetic confirms 6! / (2! * 2!) = 180 and 4! / 2! = 12, and their product is 2160. Thus the arithmetic of the complement method holds correctly.

Why Other Options Are Wrong:
The option 45360 corresponds to the total number of permutations without any restriction, which ignores the requirement that vowels are not all together. The value 216 refers to a much smaller count and does not arise naturally from the factorials used here. The value 1260 is another incorrect mixture of factorial terms. Only 43200 equals 45360 minus 2160 and correctly represents the number of arrangements where the four vowels do not form one single block.

Common Pitfalls:
Common mistakes include forgetting to account for repeated letters, which leads students to use 9! instead of 9! divided by the repetition factorials. Others mistakenly count arrangements where the vowels are together instead of not together. Some also misinterpret the condition and try to forbid any two vowels from being adjacent rather than all four. Using the complement method and carefully respecting repeated letters avoids these errors. Always pay attention to the words together and not together in such questions.

Final Answer:
The number of different arrangements of COMMITTEE in which the four vowels do not all appear together is 43200.