In how many different ways can the letters of the word CIRCUMSTANCES be arranged if each vowel (A, E, I, U) occupies an odd position and the letter N is fixed at the last (13th) position?

Introduction / Context:
This question blends permutations with multiple constraints: fixed positions, repeated consonants and vowel placement restrictions. It is a good example of how to systematically handle non trivial word arrangement problems using combinatorics, especially when some letters must appear only at certain positions and there are repeated letters.

Given Data / Assumptions:

• The word CIRCUMSTANCES has 13 letters.
• Letter counts: C appears 3 times, S appears 2 times, and I, R, U, M, T, A, N, E each appear once.
• Vowels are A, E, I and U (4 vowels).
• All vowels must be in odd positions.
• The letter N must always be at the last position, which is the 13th (an odd position).
• The remaining consonants fill the remaining positions.

Concept / Approach:
Positions 1 to 13 have 7 odd places: 1, 3, 5, 7, 9, 11 and 13. N is fixed at position 13, leaving 6 odd positions (1, 3, 5, 7, 9, 11) where the 4 distinct vowels must be placed. The 8 remaining consonant letters (C, C, C, S, S, R, M, T) then occupy the remaining 8 positions (2, 4, 6, 8, 10, 12 and the 2 unused odd positions). We count choices for vowel positions, permutations of vowels and permutations of consonants with repetition, then multiply all factors.

Step-by-Step Solution:
Odd positions available for vowels (excluding position 13): {1, 3, 5, 7, 9, 11}.Choose 4 of these 6 odd positions for the 4 vowels: number of ways = 6C4 = 15.Arrange the 4 distinct vowels (A, E, I, U) in the chosen positions: 4! = 24 ways.Remaining letters are consonants: C (3), S (2), R (1), M (1), T (1); total 8 consonants.Arrange these 8 letters in the remaining 8 positions. Because C repeats 3 times and S repeats 2 times, permutations = 8! / (3! * 2!).Compute 8! = 40320, 3! = 6 and 2! = 2, so denominator = 12, giving 40320 / 12 = 3360.Total arrangements = 15 * 24 * 3360.First, 15 * 24 = 360. Then 360 * 3360 = 1209600.

Verification / Alternative check:
You can re check multiplication: 3360 * 300 = 1008000 and 3360 * 60 = 201600, so sum = 1209600.There is no double counting because vowel positions and consonant positions are distinct and every step handles all symmetry from repeated letters.

Why Other Options Are Wrong:
151200 and 72000 are much smaller and typically arise if you ignore some positional choices or forget the division by 3! and 2! for repeated consonants.504020 does not align with any logically factored product related to this setup.

Common Pitfalls:
Placing vowels into all 7 odd positions instead of just 4, which does not reflect the actual number of vowels.Forgetting that N is fixed at the last position and accidentally treating it as a freely placeable consonant.Not accounting for the repeated consonants C and S, thereby overcounting arrangements.

Final Answer:
The required number of arrangements is 1209600.