Number of different arrangements of the word HALLUCINATION when all consonants must appear together as one group

Introduction / Context:
This problem involves counting arrangements of a long word with repeated letters and a special condition that all consonants must appear together as a single block. It combines several concepts: identification of vowels and consonants, handling repeated letters, and the block method.

Given Data / Assumptions:

• Word: HALLUCINATION.
• Letters and counts: A (2), L (2), I (2), N (2), H (1), U (1), C (1), T (1), O (1).
• Total letters: 12.
• Vowels: A, A, U, I, I, O (6 letters).
• Consonants: H, L, L, C, N, N, T (7 letters).
• All consonants must appear together as one block.

Concept / Approach:
First, treat all 7 consonants as a single block. Then arrange this block together with the 6 vowel letters, taking care of repeated letters. Next, count the internal arrangements of the consonant block itself, again handling repetitions. Finally, multiply the two counts.

Step-by-Step Solution:
Step 1: Form a consonant block C that contains H, L, L, C, N, N, T.Step 2: Outside this block, we have 6 vowels A, A, U, I, I, O.Step 3: So we now have 7 items to arrange: C and the 6 vowel letters.Step 4: Among these 7 items, vowels A and I are repeated twice each, while C is distinct as a block.Step 5: Number of ways to arrange C, A, A, U, I, I, O = 7! / (2! * 2!) = 5040 / 4 = 1260.Step 6: Now consider the internal arrangements of the consonant block C which has H, L, L, C, N, N, T.Step 7: This is 7 letters with L repeated twice and N repeated twice, so arrangements = 7! / (2! * 2!).Step 8: Compute 7! / (2! * 2!) = 5040 / 4 = 1260.Step 9: Total arrangements with all consonants together = 1260 * 1260 = 1587600.

Verification / Alternative check:
You can sanity check the magnitude: without any restriction, total arrangements of 12 letters with these repetitions would be 12! divided by the product of factorials of the repeated counts. The answer 1587600 is less than that total, as it should be, because we are imposing a restriction. The answer also matches 1260 squared, confirming the two stage block reasoning.

Why Other Options Are Wrong:

• 129780 and 35600 are much smaller than the correct count and arise from partial counting that misses either the internal or external arrangements.
• None of these is incorrect because option 1587600 already matches the correct computation.

Common Pitfalls:
Students often ignore repetition and simply use 7! or 12! leading to large incorrect numbers. Another mistake is to count only the arrangement of the block with vowels and forget to multiply by the internal arrangements of consonants. Always identify repeated letters clearly and divide by their factorials in every permutation stage.

Final Answer:
The number of valid arrangements is 1587600, so the correct answer is 1587600.