Introduction to Permutations

Introduction to Permutations

  • Definition of Permutation: A permutation is an arrangement of items where the order matters.

    • Characteristics:

    • Ordered list.

    • No repetition of items.

  • Comparison with Combination: A combination is an arrangement where the order does not matter and is meant for grouping with no repetition.

Example 1: Five Children on a Swing

  • Scenario: Five children taking turns using a swing.

  • Question: How many ways can they line up for their turns?

  • Conclusion: This is a permutation since the order of the lineup matters.

  • Notation: 5 permute 5 is denoted as <em>5P</em>5{}<em>5P</em>5 or 5P5{}^5P_5.

  • Formula for Permutation:

    • nPr=n!(n−r)!nP_r = \frac{n!}{(n - r)!}

  • Specifics for This Example:

    • Both n and r are equal to 5.

    • Calculation:

    • 5P5=5!(5−5)!=5!0!5P5 = \frac{5!}{(5 - 5)!} = \frac{5!}{0!}

    • Since 0!=10! = 1, it follows that

    • 5P5=5!5P5 = 5!

  • Calculating 5!:

    • 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120

  • Result: There are 120 ways the five children can line up to use the swing.

  • Verification using Calculator:

    • Entered as 5P5{}^5P_5 or as 5!5!.

Example 2: Tony's Summer Reading List

  • Scenario: Tony has 14 books and wants to choose 4 to read in order.

  • Question: In how many ways can he select and order these 4 books?

  • Conclusion: This is a permutation as the order matters (first to last).

  • Notation: 14 permute 4, or 14P4{}^14P_4.

  • Specifics:

    • n = 14, r = 4

    • Calculation:

    • 14P4=14!(14−4)!=14!10!14P4 = \frac{14!}{(14 - 4)!} = \frac{14!}{10!}

  • Expanding 14!:

    • 14!=14×13×12×11×10!14! = 14 \times 13 \times 12 \times 11 \times 10!

  • Simplification:

    • 14!10!=14×13×12×11\frac{14!}{10!} = 14 \times 13 \times 12 \times 11

  • Calculating the Product:

    • 14×13×12×11=24,02414 \times 13 \times 12 \times 11 = 24,024

  • Verification using Calculator:

    • Entered as 14P4{}^14P_4.

Example 3: Student Council Positions

  • Scenario: 11 students need to hold four different positions (president, vice president, secretary, treasurer).

  • Question: In how many ways can these positions be appointed?

  • Conclusion: A permutation because the order matters and no one can hold more than one position.

  • Notation: 11 permute 4, or 11P4{}^11P_4.

  • Specifics:

    • n = 11, r = 4

    • Calculation:

    • 11P4=11!(11−4)!=11!7!11P4 = \frac{11!}{(11 - 4)!} = \frac{11!}{7!}

  • Expanding 11!:

    • 11!=11×10×9×8×7!11! = 11 \times 10 \times 9 \times 8 \times 7!

  • Simplification:

    • 11!7!=11×10×9×8\frac{11!}{7!} = 11 \times 10 \times 9 \times 8

  • Calculating the Product:

    • 11×10×9×8=7,92011 \times 10 \times 9 \times 8 = 7,920

  • Verification using Calculator:

    • Entered as 11P4{}^11P_4.

Example 4: Zach's Seven-Digit Passcode

  • Scenario: Zach is creating a 7-digit passcode using digits 0-9.

  • Question: How many unique 7-digit passcodes can he create if numbers cannot be repeated?

  • Conclusion: A permutation since order matters and numbers cannot repeat.

  • Notation: 10 permute 7, or 10P7{}^10P_7.

  • Specifics:

    • n = 10, r = 7

    • Calculation:

    • 10P7=10!(10−7)!=10!3!10P7 = \frac{10!}{(10 - 7)!} = \frac{10!}{3!}

  • Expanding 10!:

    • 10!=10×9×8×7×6×5×4×3!10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!

  • Simplification:

    • 10!3!=10×9×8×7×6×5×4\frac{10!}{3!} = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4

  • Calculating the Product:

    • 10×9×8×7×6×5×4=604,80010 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 604,800

  • Verification using Calculator:

    • Entered as 10P7{}^10P_7.

Conclusion

  • The examples illustrate how to apply permutations in different contexts where both order matters and repetitions are not allowed. The essential formula for calculating permutations is crucial to evaluate such scenarios effectively.