Permutations with Repetition

[MathDefinition title="Permutations with Repetition Allowed"] When objects can be repeated, the number of permutations of n distinct objects taken r at a time is nʳ (each position has n choices). [/MathDefinition]

[MathDefinition title="Permutations of Non-Distinct Objects"] When some objects are identical, the number of distinct permutations of n objects where p₁ are of type 1, p₂ are of type 2, etc., is: n!/(p₁! × p₂! × ... × pₖ!) [/MathDefinition]

[MathKeyFormula title="Key Formulas"]

1. Permutations with repetition: nʳ
2. Permutations of identical objects: n!/(p₁! × p₂! × ... × pₖ!)
3. Permutations with restricted repetition:
   • Each object used at most once: nPr
   • Each object used exactly r times: nʳ
   • Each object used at most r times: complicated formula [/MathKeyFormula]

[MathNote title="When to Use Each Formula"]

• nʳ: When choosing r items from n types with replacement
• n!/(p₁! × p₂! × ...): When arranging objects where some are identical
• nPr: When choosing r items from n without replacement [/MathNote]

[MathExample title="Example 1: Forming Numbers with Repetition"] Problem: How many 3-digit numbers can be formed using digits 0-9 with repetition allowed?

Solution: Here n = 10 (digits 0-9), r = 3 (digits in number)

But the first digit cannot be 0: First digit: 9 choices (1-9) Second digit: 10 choices (0-9) Third digit: 10 choices (0-9)

Total numbers = 9 × 10 × 10 = 900

Alternatively, without restriction: 10³ = 1000 Numbers with 0 as first digit: 10² = 100 Valid numbers = 1000 − 100 = 900

Therefore, 900 such numbers can be formed. [/MathExample]

[MathExample title="Example 2: Arranging Letters"] Problem: Find the number of distinct permutations of the letters of the word 'MISSISSIPPI'.

Solution: MISSISSIPPI has 11 letters: M: 1, I: 4, S: 4, P: 2

Total permutations = 11!/(1! × 4! × 4! × 2!)

11! = 39916800 1! = 1 4! = 24 2! = 2

= 39916800/(1 × 24 × 24 × 2) = 39916800/1152 = 34650

Therefore, there are 34,650 distinct permutations. [/MathExample]

[MathExample title="Example 3: Forming Words with Repetition"] Problem: How many 5-letter words can be formed from the English alphabet if repetition is allowed?

Solution: Here n = 26 (English letters), r = 5 (letters in word)

Since repetition is allowed: Total words = 26⁵ = 26 × 26 × 26 × 26 × 26 = 11,881,376

Therefore, 11,881,376 such words can be formed. [/MathExample]

[MathExample title="Example 4: Distributing Objects"] Problem: In how many ways can 5 different prizes be distributed among 3 students if each student can receive any number of prizes?

Solution: Each prize can be given to any of the 3 students.

Prize 1: 3 choices Prize 2: 3 choices Prize 3: 3 choices Prize 4: 3 choices Prize 5: 3 choices

Total ways = 3⁵ = 243

Therefore, there are 243 ways to distribute the prizes. [/MathExample]