The letters of the word "CASTIGATION" are arranged at random in all possible distinct ways. What is the probability that all the vowels in the word occupy only the even positions of the arrangement?

Introduction / Context:
This is a challenging probability problem involving permutations with repeated letters and a positional restriction on vowels. You must consider all distinct arrangements of the letters in the word CASTIGATION and then determine the fraction of those arrangements in which every vowel is placed only in even positions. This problem tests your understanding of permutations with repetition and probability as a ratio of favorable to total arrangements.

Given Data / Assumptions:

• The word is CASTIGATION, which has 11 letters.
• Letters and their frequencies: C (1), A (2), S (1), T (2), I (2), G (1), O (1), N (1).
• Vowels in the word: A, A, I, I, O (total 5 vowels).
• Consonants in the word: C, S, T, T, G, N (total 6 consonants).
• We count positions from left to right as 1 to 11; even positions are 2, 4, 6, 8, 10.
• We want every vowel to appear only in positions 2, 4, 6, 8, and 10, and no consonant in those positions.

Concept / Approach:
First, compute the total number of distinct permutations of the 11 letter multiset. Then count how many of these permutations satisfy the vowel position condition. Finally, express the probability as the ratio favorable / total. Because there are repeated letters (A, T, I appearing twice), we must divide by factorials for repeated letters when counting permutations.

Step-by-Step Solution:
Step 1: Compute the total number of distinct permutations of the letters in CASTIGATION. Total letters = 11. Repeated letters: A appears 2 times, T appears 2 times, I appears 2 times. Total permutations = 11! / (2! * 2! * 2!). Step 2: Identify positions for vowels and consonants. There are 5 even positions (2, 4, 6, 8, 10) and 6 odd positions (1, 3, 5, 7, 9, 11). We must place the 5 vowels A, A, I, I, O in the 5 even positions, and the 6 consonants C, S, T, T, G, N in the 6 odd positions. Step 3: Count favorable permutations of vowels among the 5 even positions. Number of arrangements of 5 vowels (A, A, I, I, O) = 5! / (2! * 2!). Step 4: Count favorable permutations of consonants among the 6 odd positions. Number of arrangements of 6 consonants (C, S, T, T, G, N) = 6! / 2!. Step 5: Multiply the vowel and consonant arrangements to get total favorable outcomes. Favorable permutations = [5! / (2! * 2!)] * [6! / 2!]. Step 6: Now form the probability ratio. Probability = { [5! / (2! * 2!)] * [6! / 2!] } / { 11! / (2! * 2! * 2!) }. Step 7: Simplify the expression by cancelling common factors of 2! and handling factorials carefully. Carrying out the arithmetic gives a simplified fraction of 1 / 462.

Verification / Alternative check:
To verify, compute numerical values stepwise. First, 5! / (2! * 2!) counts vowel arrangements and equals 30. Next, 6! / 2! counts consonant arrangements and equals 360. Thus favorable permutations = 30 * 360 = 10800. The total permutations 11! / (2! * 2! * 2!) equal 4989600. The ratio 10800 / 4989600 can be simplified by dividing numerator and denominator by 10800, which gives 1 / 462. This is a very small probability, which makes sense because requiring all vowels to sit only in specific even positions is a strong restriction.

Why Other Options Are Wrong:
Options 1/231, 1/154, 1/330, and 1/500 are different small fractions that do not match the exact count of favorable versus total permutations. They could arise from mistakes such as miscounting the number of positions, treating some repeated letters as distinct, or failing to separate the vowel and consonant arrangements. None of these alternative fractions simplify to the correct value of 1/462.

Common Pitfalls:
Frequent mistakes include ignoring repeated letters and using 11! as the total number of permutations without dividing by the appropriate factorials for duplicates. Another issue is misidentifying the number of vowels or the number of even positions in an 11 letter word. Some learners also forget to restrict consonants entirely to odd positions, which leads to overcounting. To avoid these errors, first list the frequency of each letter, then divide positions into vowel and consonant slots, and finally compute permutations for each group separately before taking their product.

Final Answer:
The probability that all vowels in CASTIGATION occupy only the even positions is 1/462.