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

Introduction:
This is a challenging permutations and probability question involving repeated letters. You are asked to find the probability that all vowels of the word "CASTIGATION" appear only in the even positions of a randomly arranged permutation of its letters. The difficulty arises because some letters (including vowels and consonants) are repeated, so we must use permutations of multisets rather than simple factorial counts.

Given Data / Assumptions:

• The word is "CASTIGATION".
• Total letters = 11.
• Vowels: A, A, I, I, O → 5 vowels (A and I each repeated twice, O once).
• Consonants: C, S, T, G, T, N → 6 consonants (T repeated twice).
• Even positions in an 11-letter word are positions 2, 4, 6, 8 and 10 (5 positions).
• We want the probability that all vowels occupy these even positions and consonants occupy the odd positions.

Concept / Approach:
The probability is (number of favourable arrangements) divided by (total number of distinct arrangements). Because some letters repeat, we count permutations using factorials divided by factorials of repeats. For a valid arrangement, the 5 vowels must fill the 5 even spots, and the 6 consonants must fill the 6 odd spots. We count arrangements of vowels and consonants separately and then divide by the total permutations of the 11-letter multiset.

Step-by-Step Solution:
Step 1: Total number of distinct permutations of "CASTIGATION".There are 11 letters, with A repeated twice, I repeated twice, and T repeated twice.Total permutations = 11! / (2! * 2! * 2!).Step 2: Count favourable permutations.Even positions (2, 4, 6, 8, 10) must contain the 5 vowels A, A, I, I, O.Number of ways to arrange vowels = 5! / (2! * 2!) (because A and I each repeat twice).Odd positions (1, 3, 5, 7, 9, 11) must contain the 6 consonants C, S, T, G, T, N.Number of ways to arrange consonants = 6! / 2! (because T repeats twice).Total favourable arrangements = [5! / (2! * 2!)] * [6! / 2!].Step 3: Form the probability.P = favourable / total = { [5! / (2! * 2!)] * [6! / 2!] } / [11! / (2! * 2! * 2!) ].Notice that 2! * 2! * 2! cancels, leaving:P = (5! * 6!) / 11!.Compute: 5! = 120, 6! = 720, so numerator = 86400.11! = 39916800.So P = 86400 / 39916800.Simplify this fraction: divide numerator and denominator step by step to get P = 1/462.

Verification / Alternative check:
We can simplify algebraically before plugging numbers: P = (5! * 6!) / 11! = (1 * 2 * 3 * 4 * 5 * 1 * 2 * 3 * 4 * 5 * 6) / (1 * 2 * ... * 11). Cancelling common factors systematically leads to 1/462. Numerical simplification of 86400/39916800 using common divisors also confirms that the reduced fraction is 1/462 ≈ 0.00216, which is a small but reasonable probability.

Why Other Options Are Wrong:
1/231, 1/154, 1/308 and 1/420 are larger probabilities that would correspond to fewer total permutations or more favourable arrangements than actually exist. Only 1/462 matches the exact combinatorial counts derived from the repeated-letter permutations.

Common Pitfalls:
Typical mistakes include ignoring repeated letters and using plain 11! instead of 11!/(2! * 2! * 2!), or treating vowels and consonants as all distinct. Another pitfall is to misinterpret the requirement and allow vowels in odd positions. Careful counting of both vowels and consonants with their repetitions is essential to get the correct probability.

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