Permutation Formula and Problems

[MathDefinition title="Permutation"] A permutation is an arrangement of objects in a definite order. The number of permutations of n distinct objects taken r at a time is denoted by P(n,r) or ⁿPᵣ. [/MathDefinition]

[MathDefinition title="Permutation Formula"] The formula for permutations is: ⁿPᵣ = n!/(n−r)! where n! = n × (n−1) × (n−2) × ... × 2 × 1 [/MathDefinition]

[MathKeyFormula title="Key Permutation Formulas"]

1. ⁿPᵣ = n!/(n−r)!
2. ⁿPₙ = n! (arranging all n objects)
3. ⁿP₁ = n (choosing 1 from n)
4. 0! = 1 (by definition)
5. ⁿPᵣ = n × (n−1) × (n−2) × ... × (n−r+1) [r factors] [/MathKeyFormula]

[MathNote title="When to Use Permutations"] Use permutations when:

1. Order matters
2. Arranging objects in a line, circle, or specific positions
3. Selecting and arranging r objects from n distinct objects
4. Forming numbers, words, or codes where position matters [/MathNote]

[MathExample title="Example 1: Basic Permutation"] Problem: Find the number of ways to arrange 5 books on a shelf.

Solution: Here n = 5 (total books), r = 5 (arranging all)

Using the formula: ⁵P₅ = 5! = 5 × 4 × 3 × 2 × 1 = 120

Therefore, there are 120 ways to arrange 5 books on a shelf. [/MathExample]

[MathExample title="Example 2: Selecting and Arranging"] Problem: How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?

Solution: Here n = 5 (digits), r = 3 (digits in number)

Using the formula: ⁵P₃ = 5!/(5−3)! = 5!/2! = (5 × 4 × 3 × 2 × 1)/(2 × 1) = 60

Therefore, 60 three-digit numbers can be formed. [/MathExample]

[MathExample title="Example 3: Arranging with Conditions"] Problem: In how many ways can 5 men and 3 women be arranged in a row such that no two women are adjacent?

Solution: Step 1: Arrange 5 men in a row Number of ways = 5! = 120

Step 2: When 5 men are arranged, there are 6 gaps (including ends): _ M₁ _ M₂ _ M₃ _ M₄ _ M₅ _

Step 3: Choose 3 gaps out of 6 for the women Number of ways = ⁶P₃ = 6!/3! = 6 × 5 × 4 = 120

Step 4: Total arrangements = 120 × 120 = 14,400

Therefore, there are 14,400 ways to arrange them. [/MathExample]

[MathExample title="Example 4: Forming Words"] Problem: How many different 4-letter words can be formed from the letters of the word 'SYSTEM' without repetition?

Solution: The word SYSTEM has 6 distinct letters: S, Y, S, T, E, M Wait, S appears twice, so we have 5 distinct letters: S, Y, T, E, M (S repeated)

Actually, SYSTEM has letters: S, Y, S, T, E, M Distinct letters: S, Y, T, E, M (5 distinct, S repeated twice)

For 4-letter words without repetition from 5 distinct letters: ⁵P₄ = 5!/(5−4)! = 5!/1! = 120

But wait, if we can use S only once (since we're choosing without repetition), and we have 5 distinct letters, then: ⁵P₄ = 5 × 4 × 3 × 2 = 120

However, if we consider that S appears twice in the original word, we might need to account for that differently. Let me clarify:

If we're selecting 4 letters from the 6 positions (S₁, Y, S₂, T, E, M) without repetition: ⁶P₄ = 6!/2! = 360

But if we're selecting 4 distinct letters from {S, Y, T, E, M}: ⁵P₄ = 120

The answer depends on interpretation. The most common interpretation is 120. [/MathExample]