Comprehensive Study Guide for Extension 1 Mathematics: Permutations and Combinations

The Multiplication Rule and Fundamental Counting

  • Definition of the Multiplication Rule: If a sequence of events can happen in multiple stages, the total number of ways the sequence can occur is found by multiplying the number of ways each individual event can occur. If the first event can occur in mm ways, the second in nn ways, and a third in rr ways, then the total number of arrangements for the three events is m×n×rm \times n \times r.

  • Example: Outfit Selection:

    • Scenario: Erin is choosing an outfit and has 5 tops, 6 skirts, and 4 caps.
    • Calculation: In how many ways can she select one top, one skirt, and one cap?
    • Solution: Ways=5×6×4\text{Ways} = 5 \times 6 \times 4

Repetition of Events and Ordered Arrangements

  • General Rule for Repetition: If an event has nn possible outcomes and occurs rr times with repetition allowed, the number of ordered arrangements is given by the formula nrn^r.

  • Example 1: Rolling a Die:

    • Rolling a die 2 times: 6×6=626 \times 6 = 6^2
    • Rolling a die 3 times: 6×6×6=636 \times 6 \times 6 = 6^3
    • Rolling a die rr times: 6×6×6×⋯=6r6 \times 6 \times 6 \times \dots = 6^r

  • Example 2: Car Number Plates:

    • (a) Total Possible Plates: A plate consists of 3 letters followed by 3 digits (repetition allowed).
      • Calculation: 26×26×26×10×10×10=263×10326 \times 26 \times 26 \times 10 \times 10 \times 10 = 26^3 \times 10^3
    • (b) Specific Sequence: How many plates begin with the specific letters "ABC"?
      • Calculation: 1×1×1×10×10×10=1031 \times 1 \times 1 \times 10 \times 10 \times 10 = 10^3
    • (c) Probability: If a plate is chosen at random, what is the probability it begins with "ABC"?
      • Solution: 103263×103=1263\frac{10^3}{26^3 \times 10^3} = \frac{1}{26^3}

Factorial Representation

  • Definition: The factorial of a positive integer nn, denoted as n!n!, is the product of all positive integers less than or equal to nn.

    • Formula: n!=n(n−1)(n−2)…3×2×1n! = n(n - 1)(n - 2) \dots 3 \times 2 \times 1
    • Example: 5!=5×4×3×2×15! = 5 \times 4 \times 3 \times 2 \times 1
    • Special Note: 0!=10! = 1

  • Examples of Factorial Applications:

    • Arranging 6 people in a row: 6×5×4×3×2×1=6!6 \times 5 \times 4 \times 3 \times 2 \times 1 = 6!
    • Choosing and arranging 3 people from 6: 6×5×4=1206 \times 5 \times 4 = 120

Arrangements or Permutations (nPrnPr)

  • Definition: Distinctly ordered sets are called arrangements or permutations. Order is the distinguishing factor.

  • The Permutation Formula: The number of permutations of nn objects taken rr at a time is given by:

    • nPr=n!(n−r)!nPr = \frac{n!}{(n - r)!}
    • nn = total number of objects.
    • rr = number of positions/selections.

  • Example 1: Debating Team (4 speakers):

    • (a) Arranging all 4 for a photo: 4×3×2×1=4!4 \times 3 \times 2 \times 1 = 4! or 4P44P4
    • (b) Choosing a captain and vice-captain: 4×3=124 \times 3 = 12 or 4P24P2

  • Example 2: Horse Racing (7 horses):

    • (a) Total finishing orders: 7×6×5×4×3×2×1=7!7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 7! or 7P77P7
    • (b) Possible trifectas (1st, 2nd, and 3rd): 7×6×5=2107 \times 6 \times 5 = 210 or 7P37P3

Permutations with Restrictions

  • Problem Scenario: Arranging 5 boys and 4 girls (9 total) on a bench.
    • a) No restrictions: 9!9! or 9P99P9
    • b) Boys and girls alternate: Since there are 5 boys and 4 girls, a boy must be on each end (BGBGBGBGB).
      • Calculation: 5×4×4×3×3×2×2×1×1=5!×4!5 \times 4 \times 4 \times 3 \times 3 \times 2 \times 2 \times 1 \times 1 = 5! \times 4! or 5P5×4P45P5 \times 4P4
    • c) Boys and girls in separate groups: Either all boys then all girls, or all girls then all boys.
      • Calculation: 5!×4!+4!×5!=5!×4!×25! \times 4! + 4! \times 5! = 5! \times 4! \times 2 or 5P5×4P4×25P5 \times 4P4 \times 2
    • d) Anne and Jim stay together: Treat (Anne and Jim) as one block.
      • Calculation: 2×8!2 \times 8! or 2×8P82 \times 8P8 (The 2 accounts for rearranging Anne and Jim within their block, and 8! for the 8 entities: the block and the other 7 individuals).

Arrangements with Repetitions (Alike Objects)

  • Formula: If there are nn elements where xx are alike of one kind, yy are alike of another kind, and zz are alike of another, the number of permutations is:

    • n!x!y!z!\frac{n!}{x! y! z!}

  • Example: PARRAMATTA:

    • The word contains 10 letters: 4 A's, 2 R's, 2 T's, 1 P, and 1 M.
    • Number of arrangements: 10!4!2!2!=37800\frac{10!}{4! 2! 2!} = 37800

Detailed Word Arrangement Examples: REMAND

  • Case study: REMAND (6 unique letters):
    • a) No restrictions: 6P6=7206P6 = 720 or 6!6!
    • b) Begin with RE: Fixed positions for R and E at the start (RE____R E \_ \_ \_ \_).
      • Calculation: 4P4=244P4 = 24 or 4!4!
    • c) Do not begin with RE: Total arrangements minus those that do begin with RE.
      • Calculation: 6!−4!=6966! - 4! = 696
    • d) RE together in that order: Treat (RE) as a single item.
      • Calculation: 5P5=1205P5 = 120 or 5!5!
    • e) REM together in any order: Treat (REM) as a single item that can be arranged in 3!3! ways.
      • Calculation: 3P3×4P4=1443P3 \times 4P4 = 144 (equivalent to 3!×4!3! \times 4!)
    • f) R, E, and M are not to be together: Total minus those where they are together.
      • Calculation: 6!−144=5766! - 144 = 576

Advanced Linear Restrictions

  • Boat Seating Example: 6 boys entering a boat with 8 seats (4 on each side).

    • (a) Sit anywhere: 8P68P6
    • (b) Specific positioning: Two boys (A and B) on the port side, one boy (W) on the starboard side.
      • Port side selection/arrangement for A and B: 4P24P2
      • Starboard side selection for W: 4P14P1
      • Arrangement for remaining 3 boys in the remaining 5 seats: 5P35P3
      • Total: 4P2×4P1×5P34P2 \times 4P1 \times 5P3

  • Numerical Restrictions (Digits 2, 3, 4, 5, 6):

    • (a) Numbers greater than 4000:
      • 5-digit numbers: 5P55P5
      • 4-digit numbers (must start with 4, 5, or 6): 3P1×4P33P1 \times 4P3
      • Total: 5P5+(3P1×4P3)5P5 + (3P1 \times 4P3)
    • (b) Even 4-digit numbers: Must end with 2, 4, or 6.
      • Calculation: End digit (3 choices)×Remaining 3 slots (from remaining 4 digits)=4P3×3P1\text{End digit (3 choices)} \times \text{Remaining 3 slots (from remaining 4 digits)} = 4P3 \times 3P1

Circular Arrangements

  • Concept: Arrangements in a circle are different because rotating the circle doesn't change the relative positions. While 5 objects (a, b, c, d, e) have distinct line arrangements (abcde, bcdea, etc.), they are identical in a circle.

  • Formula: To arrange nn objects in a circle, fix one position to break the symmetry. The remaining (n−1)(n - 1) objects are arranged as if on a line.

    • Arrangements in a circle=(n−1)!\text{Arrangements in a circle} = (n - 1)!

  • Round Table Examples (6 Men, 6 Women):

    • a) No restrictions: (12−1)!=11!(12 - 1)! = 11!
    • b) Men and women alternate: Fix one man. Arrange the other 5 men in (6−1)!(6 - 1)! ways. Arrange the 6 women in the 6 gaps in 6!6! ways.
      • Calculation: 5!×6!5! \times 6!
    • c) Ted and Carol together: Treat (TC) as one unit (2!2! ways inside). Then arrange the unit and the other 10 people in a circle.
      • Calculation: 2!×10!2! \times 10!
    • d) Bob, Ted, and Carol together: Treat (BTC) as one unit (3!3! ways inside).
      • Calculation: 3!×9!3! \times 9!
    • e) Neither Bob nor Carol sit next to Ted: Seat 2 of the other 9 people next to Ted (9P29P2). Then seat the remaining 9 people (including Bob and Carol).
      • Calculation: 9P2×9!9P2 \times 9!

  • Necklaces and Beads: When a circular arrangement can be flipped over (like a necklace), clockwise and anti-clockwise arrangements are considered identical.

    • Example: 8 differently colored beads on a string.
    • Solution: (8−1)!2=7!2\frac{(8 - 1)!}{2} = \frac{7!}{2}

Unordered Selections (Combinations)

  • Definition: Combinations refer to selections where the order does not matter.

  • The Combination Formula (nCrnCr):

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

  • Basic Examples:

    • Basketball team: Choose 5 players from 8.
      • Solution: 8C58C5

  • Committee Selection (6 Men, 4 Women):

    • a) No restrictions (5 people): 10C510C5
    • b) One particular person included: 1×9C41 \times 9C4
    • c) One particular woman excluded: 9C59C5
    • d) 3 men and 2 women: 6C3×4C26C3 \times 4C2
    • e) Men only: 6C56C5
    • f) Majority of women: Could be (3 Women and 2 Men) OR (4 Women and 1 Man).
      • Calculation: 4C3×6C2+4C4×6C14C3 \times 6C2 + 4C4 \times 6C1

Applications in Card Games (Poker)

  • Hand: 5 cards dealt from a standard 52-card pack.
    • (i) No restrictions: 52C552C5
    • (ii) Specific Hands:
      • a) 4 Kings: 4C4×48C14C4 \times 48C1 or 1×481 \times 48
      • b) 2 Clubs and 3 Hearts: 13C2×13C313C2 \times 13C3
      • c) All Hearts: 13C513C5
      • d) All the same color: (All Red OR All Black).
        • Calculation: 26C5+26C5=2×26C526C5 + 26C5 = 2 \times 26C5
      • e) Four of the same kind: There are 13 possible "kinds" (Aces through Kings).
        • Calculation: 13×(4C4×48C1)=13×1×4813 \times (4C4 \times 48C1) = 13 \times 1 \times 48
      • f) 3 Aces and 2 Kings: 4C3×4C24C3 \times 4C2

Mixed Permutation and Combination Problems

  • Book Shelf Example: Selecting 4 Maths books (from 6) and 3 English books (from 5), then arranging the 7 selected books on a shelf.
    • a) No restrictions: Selection (Maths) ×\times Selection (English) ×\times Arrangement of the 7.
      • Calculation: 6C4×5C3×7!6C4 \times 5C3 \times 7!
    • b) 4 Maths books remain together: Treat the 4 selected Maths books as one block.
      • Calculation: 6P4×5C3×4!6P4 \times 5C3 \times 4! (Note: 6P46P4 covers selecting and arranging the Maths books, 5C35C3 selects English, then 4!4! arranges the block plus 3 English books).
      • Alternative interpretation: (6C4×4!)×5C3×4!(6C4 \times 4!) \times 5C3 \times 4!
    • c) Maths book at the beginning:
      • Calculation: 6×5C3×5C3×6!6 \times 5C3 \times 5C3 \times 6! (Where 6 is the choice for the first book, and remaining slots filled by combinations/permutations).
    • d) Maths and English alternate: Pattern is M E M E M E M.
      • Calculation: 6P4×5P36P4 \times 5P3
    • e) Maths at start and English in middle: M__E___M \_ \_ E \_ \_ \_
      • Calculation: 6×5×5C3×4C2×5!6 \times 5 \times 5C3 \times 4C2 \times 5!

Probability with Identical Elements

  • Word: SYLLABUS:
    • Length: 8 letters (Includes 2 S's and 2 L's).
    • Total arrangements: 8!2!×2!=10080\frac{8!}{2! \times 2!} = 10080
    • (a) Probability two S's are together:
      • Arrangements with (SS) treated as one block: 7!2!\frac{7!}{2!} (the 2!2! accounts for the 2 L's).
      • Result: 25202520
      • Probability: 252010080=14\frac{2520}{10080} = \frac{1}{4}
    • (b) Probability begins and ends with L:
      • Format: L______LL \_ \_ \_ \_ \_ \_ L (L's are fixed).
      • Arrangements of remaining 6 letters (contains 2 S's): 6!2!\frac{6!}{2!}.
      • Result: 360360
      • Probability: 36010080=128\frac{360}{10080} = \frac{1}{28}