đź“ť Word Rearrangements

Problems involving arranging letters with repetition, position constraints, and other restrictions.

🎯 Recognition Cues

  • Keywords: “arrange letters”, “rearrange”, “anagram”, “word”, “string”
  • Setup: Letters with or without repetition
  • Constraints: Specific positions, first/last letters, adjacency, patterns

đź“‹ Solution Template

  1. Identify the letter types:

    • All distinct letters
    • Some repeated letters
    • Indistinguishable letters

  2. Apply the appropriate formula:

    • All distinct: $n!$ arrangements
    • Some repeated: $\frac{n!}{n_1!n_2!\cdots n_k!}$ (multinomial)
    • Indistinguishable: $\binom{n}{k}$ (choose positions)

  3. Apply constraints:

    • Fixed positions: Arrange the rest
    • First/last letters: Choose, then arrange
    • Adjacency: Treat as unit, then arrange

đź’ˇ Micro-Examples

Basic Rearrangement

  • Problem: How many ways can you arrange the letters in “MATH”?
  • Solution: $4! = 24$ ways (all distinct letters)

With Repetition

  • Problem: How many ways can you arrange the letters in “MISSISSIPPI”?
  • Solution: $\frac{11!}{1!4!4!2!} = 34650$ ways (M:1, I:4, S:4, P:2)

Position Constraints

  • Problem: How many ways can you arrange “MATH” so M is first?
  • Solution: Fix M in first position, arrange the rest: $3! = 6$ ways

Adjacency Constraints

  • Problem: How many ways can you arrange “MISSISSIPPI” so no two I’s are adjacent?
  • Solution: Arrange M, S, S, S, S, P, P first: $\frac{7!}{4!2!} = 105$ ways, then place I’s in 8 gaps: $\binom{8}{4} = 70$ ways. Total: $105 \cdot 70 = 7350$ ways

⚠️ Common Pitfalls & Variants

Pitfall: Forgetting to divide by factorials of repeated letters

  • Wrong: “Arrange AABBC” = $5! = 120$ ways
  • Right: $\frac{5!}{2!2!1!} = 30$ ways (divide by factorials of repeated letters)

Pitfall: Confusing arrangements with selections

  • Wrong: “How many 3-letter words from MATH?” = $4!$ ways
  • Right: $P(4,3) = 4 \cdot 3 \cdot 2 = 24$ ways (arrangements of 3 from 4)

Pitfall: Adjacency with repetition

  • Wrong: “No two I’s adjacent in MISSISSIPPI” = Arrange all letters, subtract adjacent cases
  • Right: Use gaps method - arrange non-I’s first, then place I’s in gaps

🏆 AMC-Style Worked Example

Problem: How many ways can you arrange the letters in “AMC” so that no two vowels are adjacent?

Solution:

  • Step 1: Identify vowels and consonants

    • Vowels: A (only one vowel)
    • Consonants: M, C

  • Step 2: Since there’s only one vowel, the condition is automatically satisfied

  • Step 3: Arrange all 3 letters: $3! = 6$ ways

Wait, let me reconsider…

  • Step 1: Vowels: A (only one vowel)
  • Step 2: With only one vowel, it’s impossible to have two vowels adjacent
  • Step 3: Arrange all 3 letters: $3! = 6$ ways

Key insight: Sometimes the constraint is automatically satisfied!

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