In how many different ways can the letters of the word "MATHEMATICS" be arranged so that all the vowels in the word always appear together as one block?

Introduction / Context:
This problem combines the idea of permutations with repeated letters and a block constraint on vowels. The word "MATHEMATICS" contains repeated consonants and vowels, and we are asked to treat all vowels as if they form a single unit. Such questions are common in advanced permutation topics where both grouping and repetition need to be handled correctly.

Given Data / Assumptions:

• Word: MATHEMATICS.
• Total letters = 11.
• Consonants: M, T, H, M, T, C, S (7 consonants, with M repeated twice and T repeated twice).
• Vowels: A, A, E, I (4 vowels, with A repeated twice).
• Vowels must appear together as one contiguous block anywhere in the arrangement.
• All letters of the word must be used in each arrangement.

Concept / Approach:
We use a two stage approach. First, treat the group of all vowels as a single block. This block joins the consonants, giving fewer effective objects to arrange, but some of these have repetitions. Second, count the internal arrangements of the vowels inside the block, again taking repetition into account. The total number of valid arrangements is the product of the number of ways to arrange the block with consonants and the number of ways to arrange the letters within the vowel block itself.

Step-by-Step Solution:
Step 1: Treat the vowels A, A, E, I as one single block V. Then we have the objects: V, M, A (already inside V, so not separate), T, H, E, M, A, T, I, C, S. Practically, we count 7 consonants: M, M, T, T, H, C, S, plus the block V, so total objects = 8. Among these 8 objects, M appears 2 times and T appears 2 times; V, H, C, S each appear 1 time. Number of ways to arrange these 8 objects = 8! / (2! * 2!). Compute 8! = 40320 and divide by (2 * 2) = 4 to get 40320 / 4 = 10080. Step 2: Arrange the 4 vowels A, A, E, I inside the block V. Number of internal vowel arrangements = 4! / 2! (because A repeats twice). Compute 4! = 24; divide by 2 to get 12. Total valid arrangements = 10080 * 12 = 120960.

Verification / Alternative check:
If there were no vowel grouping restriction, the total arrangements of "MATHEMATICS" would be 11! divided by 2! for the two M's, by another 2! for the two T's and by 2! for the two A's. That value is much larger than 120960, so it is reasonable that the constraint reduces the count. Our method cleanly separates the external block permutations from the internal vowel permutations, and the arithmetic is straightforward. This supports confidence in the final answer of 120960.

Why Other Options Are Wrong:

• 120000: Very close but slightly smaller, usually coming from rounding or skipping one division step.
• 146700: Too large; this often reflects miscounting the duplicated letters or missing a division by a factorial for repeats.
• 72000: Too small; it suggests that either the vowel arrangements or the duplicated consonants were mishandled.

Common Pitfalls:
Common mistakes include forgetting about repeated letters when taking factorials, treating the vowel block incorrectly as if its internal order does not matter, or double counting by multiplying extra factors. Some learners accidentally include A, A, E and I as separate objects in the outer arrangement as well as inside the block. The safe approach is to first clearly list consonants and vowels, compress vowels into one block, handle repetition in each step and then multiply the two results.

Final Answer:
The number of different arrangements of the letters of "MATHEMATICS" where all vowels appear together as one block is 120960.