In how many different permutations can the letters of the word CORPORATION be arranged so that all the vowels always come together as a single block?

Introduction / Context:
This arrangement problem asks us to treat all vowels in the word CORPORATION as a single block so that they always stay together. The word contains repeated letters, which means we must use permutations of multiset letters rather than simple factorials. Once we group the vowels into one block, we arrange that block and the consonants, taking into account the repeated letters, and then arrange the vowels inside the block.

Given Data / Assumptions:

• The word is CORPORATION.
• Letters and frequencies: C(1), O(3), R(2), P(1), A(1), T(1), I(1), N(1).
• Total letters = 11.
• Vowels: O, O, O, A, I (5 letters, O repeated 3 times).
• Consonants: C, R, R, P, T, N (6 letters, R repeated 2 times).
• All vowels must appear together in every arrangement.

Concept / Approach:
To keep all vowels together, we think of the vowels as forming a single super letter or block. Then we count permutations of this block together with the consonants. Because there are repeated consonants (two R letters), we divide by the factorial of the repetition count. After finding the number of ways to arrange the block plus consonants, we multiply by the number of ways to arrange the vowels within the block, again dividing by the factorial of the repeated vowel O. The total number of valid permutations is the product of these two factors.

Step-by-Step Solution:
Step 1: Consider all vowels as a single block V. We have block V plus consonants C, R, R, P, T, N. Total objects to arrange now = 1 (block V) + 6 consonants = 7 objects. Among these, R appears twice. Step 2: Count ways to arrange the 7 objects with repeated R. Number of permutations of these 7 objects = 7! / 2!. 7! = 5040, and 2! = 2, so 7! / 2! = 2520. Step 3: Count ways to arrange vowels inside block V. Vowels: O, O, O, A, I (5 letters with O repeated 3 times). Number of permutations = 5! / 3!. 5! = 120 and 3! = 6, so 5! / 3! = 120 / 6 = 20. Step 4: Multiply these results. Total valid permutations = (7! / 2!) * (5! / 3!) = 2520 * 20 = 50400.

Verification / Alternative check:
We can check the logic by considering that if there were no repetition, we would simply take 7! for the block and consonants and 5! for the vowels. The repetitions of R and O require division by 2! and 3! respectively, which is exactly what we have done. No other letters are repeated, so no further divisions are needed. The final product 50400 is consistent with a multiset permutation approach.

Why Other Options Are Wrong:
The value 50000 is close but does not follow from any factorial expression with the correct repetitions. The value 40500 is also not equal to 2520 * 20. The value 5040 corresponds to 7! with no adjustments and ignores vowel arrangements and repetitions. The value 72000 is larger than the correct total and suggests that repetition handling or block arrangement has been misapplied.

Common Pitfalls:
Students often forget to treat the vowels as a single block initially or ignore some of the repeated letters. Another error is to multiply by 5! instead of 5! / 3!, thereby overcounting arrangements of vowels due to the identical O letters. Similarly, some learners forget to divide by 2! for the repeated R. Always list letter frequencies, treat the vowel block as a single object first, and apply multiset permutation formulas carefully.

Final Answer:
The letters of the word CORPORATION can be arranged with all vowels together in 50400 different ways.