Chapter 05: Probability - Permutation, Combination, and Binomial Expansion

Factorial Notation

  • Definition of Factorial: The notation n!n! is read as "n factorial" and represents the product of the first nn natural numbers. It is denoted as the product 1×2×3×4×⋯×n1 \times 2 \times 3 \times 4 \times \dots \times n.

  • Conceptual Representation:

    • 1=1!1 = 1!

    • 1×2=2!1 \times 2 = 2!

    • 1×2×3=3!1 \times 2 \times 3 = 3!

    • 1×2×3×4=4!1 \times 2 \times 3 \times 4 = 4!

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

  • Important Definition for Zero: By definition, 0!=10! = 1.

  • Example 1: Simplify Factorials:

    • a. 7!=7×6×5×4×3×2×1=50407! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040

    • b. 3!×4!=(3×2×1)×(4×3×2×1)=6×24=1443! \times 4! = (3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1) = 6 \times 24 = 144

    • c. 7!5!=7×6×5!5!=42\frac{7!}{5!} = \frac{7 \times 6 \times 5!}{5!} = 42

    • d. 12!10!×2!=12×11×10!10!×2×1=1322=66\frac{12!}{10! \times 2!} = \frac{12 \times 11 \times 10!}{10! \times 2 \times 1} = \frac{132}{2} = 66

  • Exercise 1 Tasks:

    • Simplify: 8!8!

    • Simplify: 10!7!\frac{10!}{7!}

    • Simplify: 4!3!\frac{4!}{3!}

Permutations

  • Definition: A permutation is the choice of rr things from a set of nn things without replacement where the order of selection matters.

  • Permutation Formula:     P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}

  • Numerical Examples of Permutations:

    • a. P(7,2)=7!(7−2)!=7!5!=7×6×5!5!=42P(7, 2) = \frac{7!}{(7 - 2)!} = \frac{7!}{5!} = \frac{7 \times 6 \times 5!}{5!} = 42

    • b. P(10,3)=10!(10−3)!=10!7!=10×9×8×7!7!=720P(10, 3) = \frac{10!}{(10 - 3)!} = \frac{10!}{7!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 720

    • c. P(8,5)P(8, 5) (listed as P(8,3)P(8, 3) in the simplification step): 8!(8−5)!=8!3!=8×7×6×5×4×3!3!=6720\frac{8!}{(8 - 5)!} = \frac{8!}{3!} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3!} = 6720

  • Arrangement Examples:

    • Arranging all letters in the word EARTH: The word consists of 5 distinct letters. Total ways = 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120

    • Arranging 3 books on a shelf of 5: This is a permutation of 5 objects taken 3 at a time (P(5,3)P(5, 3)). P(5,3)=5!(5−3)!=5!2!=5×4×3×2!2!=60P(5, 3) = \frac{5!}{(5 - 3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2!}{2!} = 60

    • Arranging three letters from the word HISTORY: The word HISTORY has 7 distinct letters. Total ways = P(7,3)=7!(7−3)!=7!4!=7×6×5×4!4!=210P(7, 3) = \frac{7!}{(7 - 3)!} = \frac{7!}{4!} = \frac{7 \times 6 \times 5 \times 4!}{4!} = 210

  • Exercise 2 Tasks:

    1. Find: P(6,2)P(6, 2), P(7,4)P(7, 4), and P(10,3)P(10, 3).

    2. How many ways can you arrange all letters in the word PRESIDENT?

    3. How many ways can you arrange four letters in the word FATHER?

    4. How many words (with or without meaning) can be formed using all letters of the word EQUATION using each letter exactly once?

Permutations with Identical Objects (Non-distinct Objects)

  • Theorem of Permutation: The number of permutations of nn objects where p1p_1 are of one kind, p2p_2 of another kind, …, and pkp_k of another kind is given by the formula:     n!p1!p2!…pk!\frac{n!}{p_1! p_2! \dots p_k!}

  • Example 3 Arranging Words:

    • a. HAAB-HAABAD: Total letters = 10. The letters 'A' repeat 5 times, 'H' repeats 2 times, and 'B' repeats 2 times.         Ways=10!5!2!2!=10×9×8×7×6×5!5!×2×1×2×1=302404=7560\text{Ways} = \frac{10!}{5! 2! 2!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 2 \times 1 \times 2 \times 1} = \frac{30240}{4} = 7560

    • b. TEST: Total letters = 4. The letter 'T' repeats 2 times.         Ways=4!2!=4×3×2!2!=12\text{Ways} = \frac{4!}{2!} = \frac{4 \times 3 \times 2!}{2!} = 12

  • Exercise 3 Tasks: Find the number of permutations for the words:

    • GABOOBAY

    • DUUFAAN

    • MATHEMATICS

Permutations with Repetition and Constraints

  • Example 1: 4-digit numbers from the set {1, 2, 3, 4, 5, 6}:

    • a. Repetition IS allowed: Each of the 4 slots has 6 choices.         6×6×6×6=1296 ways6 \times 6 \times 6 \times 6 = 1296 \text{ ways}

    • b. Repetition is NOT allowed: Permutation of 6 things taken 4 at a time (P(6,4)P(6, 4)).         6×5×4×3=360 ways6 \times 5 \times 4 \times 3 = 360 \text{ ways}

  • Example 2: 3-digit numbers from the set {4, 5, 6, 7, 8}:

    • a. Repetition IS allowed: 5×5×5=125 ways5 \times 5 \times 5 = 125 \text{ ways}

    • b. Repetition is NOT allowed: 5×4×3=60 ways5 \times 4 \times 3 = 60 \text{ ways}

  • Example 3: 3-digit numbers greater than 200 from the set {1, 2, 3, 4, 5}:

    • a. Repetition IS allowed: The first digit must be 2, 3, 4, or 5 (4 ways). The subsequent digits can be any of the 5.         4×5×5=100 ways4 \times 5 \times 5 = 100 \text{ ways}

    • b. Repetition is NOT allowed: The first digit has 4 ways (2, 3, 4, or 5). The second digit has 4 remaining ways, and the third has 3.         4×4×3=48 ways4 \times 4 \times 3 = 48 \text{ ways}

  • Example 4: 3-digit odd numbers from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}:

    • a. Repetition IS allowed: 5×5×5=125 ways5 \times 5 \times 5 = 125 \text{ ways}

    • b. Repetition is NOT allowed: 5×4×3=60 ways5 \times 4 \times 3 = 60 \text{ ways}     (Note: Per the transcript, these values are used even though standard logic might use the full set for non-restricted slots).

  • Exercise 4 & 5 Tasks:

    1. 4-digit numbers from {3, 4, 5, 6, 7, 8} (allowed vs not allowed).

    2. 2-digit numbers from {6, 7, 8} (allowed vs not allowed).

    3. 4-digit numbers greater than 5000 from {4, 5, 6, 7, 8}.

    4. 3-digit numbers greater than 200 from {1, 2, 3, 4, 5, 6}.

    5. 7-digit telephone numbers from {0-9} where the first three digits are 471.

    6. 5-digit telephone numbers starting with 67 with no digit appearing more than once.

Solving Equations involving Permutations

  • Solving for r in P(5,r)=60P(5, r) = 60:     5!(5−r)!=60  ⟹  120(5−r)!=60  ⟹  (5−r)!=2\frac{5!}{(5 - r)!} = 60 \implies \frac{120}{(5 - r)!} = 60 \implies (5 - r)! = 2     Since 2!=22! = 2, then 5−r=2  ⟹  r=35 - r = 2 \implies r = 3.

  • Solving for n in P(n,2)=42P(n, 2) = 42:     n!(n−2)!=42  ⟹  n(n−1)=42  ⟹  n2−n−42=0\frac{n!}{(n - 2)!} = 42 \implies n(n - 1) = 42 \implies n^2 - n - 42 = 0     Factoring: (n−7)(n+6)=0(n - 7)(n + 6) = 0.     Since nn cannot be negative, n=7n = 7.

  • Complex Identity Calculation:     Solve for nn: P(n,5)=42×P(n,3)P(n, 5) = 42 \times P(n, 3)     n!(n−5)!=42×n!(n−3)!\frac{n!}{(n - 5)!} = 42 \times \frac{n!}{(n - 3)!}     1=42×(n−5)!(n−3)(n−4)(n−5)!1 = 42 \times \frac{(n - 5)!}{(n - 3)(n - 4)(n - 5)!}     (n−3)(n−4)=42  ⟹  n2−7n+12=42  ⟹  n2−7n−30=0(n - 3)(n - 4) = 42 \implies n^2 - 7n + 12 = 42 \implies n^2 - 7n - 30 = 0     (n−10)(n+3)=0  ⟹  n=10(n - 10)(n + 3) = 0 \implies n = 10 (since n > 4).

  • Verifying an Identity:     P(n,4)−P(n−1,4)=4×P(n−1,3)P(n, 4) - P(n - 1, 4) = 4 \times P(n - 1, 3)     Example with n=5n=5:     P(5,4)−P(4,4)=120−24=96P(5, 4) - P(4, 4) = 120 - 24 = 96     4×P(4,3)=4×24=964 \times P(4, 3) = 4 \times 24 = 96     96=9696 = 96 (Identity verified).

  • Exercise 6 & 7 Tasks:

    • Solve P(12,r)=1320P(12, r) = 1320

    • Solve P(n,4)=360P(n, 4) = 360

    • Solve P(7,r)=210P(7, r) = 210

    • Solve P(n,8)=2×P(n,6)P(n, 8) = 2 \times P(n, 6)

    • Solve P(n−1,3):P(n,4)=1:9P(n - 1, 3) : P(n, 4) = 1 : 9

Combinations

  • Definition: A combination is a selection of items from a set where the order of selection does not matter.

  • Combination Formula:     C(n,r)=(nr)=n!r!(n−r)!C(n, r) = \binom{n}{r} = \frac{n!}{r!(n - r)!}

  • Example 1: Evaluate Combinations:

    • a. C(5,2)=5!2!(5−2)!=5!2!3!=5×4×3!2×1×3!=202=10C(5, 2) = \frac{5!}{2!(5 - 2)!} = \frac{5!}{2! 3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10

    • b. C(7,4)=7!4!(7−4)!=7!4!3!=7×6×5×4!4!×3×2×1=2106=35C(7, 4) = \frac{7!}{4!(7 - 4)!} = \frac{7!}{4! 3!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} = \frac{210}{6} = 35

  • Word Problems with Combinations:

    • Reading books: Reading 5 books from a list of 8 where order doesn't matter:         C(8,5)=8!5!3!=8×7×6×5!5!×6=56 waysC(8, 5) = \frac{8!}{5! 3!} = \frac{8 \times 7 \times 6 \times 5!}{5! \times 6} = 56 \text{ ways}

    • Committee selection: From 2 men and 3 women, select a committee of 3.         Total ways: C(5,3)=5!3!2!=10C(5, 3) = \frac{5!}{3! 2!} = 10         Ways to have 1 man and 2 women: C(2,1)×C(3,2)=2×3=6 waysC(2, 1) \times C(3, 2) = 2 \times 3 = 6 \text{ ways}

    • Ball selection: Bag contains 5 black and 6 red balls. Select 2 black and 2 red.         C(5,2)×C(6,2)=10×15=150 waysC(5, 2) \times C(6, 2) = 10 \times 15 = 150 \text{ ways}

  • Exercise 8 Tasks:

    1. Find: C(6,2)C(6, 2), C(8,5)C(8, 5), C(7,4)C(7, 4).

    2. Team of 3 boys and 3 girls from 5 boys and 4 girls.

    3. Selecting 9 balls from 6 red, 5 white, and 5 blue (3 of each color).

Binomial Expansion

  • Definition: The sum (a+b)(a + b) is a binomial. Any expansion of (a+b)n(a + b)^n is the power of a binomial.

  • General Principle of Expansion:

    • (a+b)0=1(a + b)^0 = 1

    • (a+b)1=a+b(a + b)^1 = a + b

    • (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

    • (a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

    • (a+b)4=a4+4a3b+6a2b2+4ab3+b4(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

  • Example 1: Use (a+b)3(a + b)^3 expansion:

    • a. (x+3)3(x + 3)^3:         (x)3(3)0+3(x)2(3)1+3(x)1(3)2+(x)0(3)3=x3+9x2+27x+27(x)^3(3)^0 + 3(x)^2(3)^1 + 3(x)^1(3)^2 + (x)^0(3)^3 = x^3 + 9x^2 + 27x + 27

    • b. (2x+4)3(2x + 4)^3:         (2x)3+3(2x)2(4)+3(2x)(4)2+(4)3=8x3+3(4x2)(4)+3(2x)(16)+64(2x)^3 + 3(2x)^2(4) + 3(2x)(4)^2 + (4)^3 = 8x^3 + 3(4x^2)(4) + 3(2x)(16) + 64         =8x3+48x2+96x+64= 8x^3 + 48x^2 + 96x + 64

  • Example 2: Use (a+b)4(a + b)^4 expansion:

    • a. (x2+2)4(x^2 + 2)^4:         (x2)4+4(x2)3(2)+6(x2)2(2)2+4(x2)(2)3+(2)4(x^2)^4 + 4(x^2)^3(2) + 6(x^2)^2(2)^2 + 4(x^2)(2)^3 + (2)^4         =x8+8x6+24x4+32x2+16= x^8 + 8x^6 + 24x^4 + 32x^2 + 16

    • b. (x−4)4(x - 4)^4:         (x)4+4(x)3(−4)+6(x)2(−4)2+4(x)(−4)3+(−4)4(x)^4 + 4(x)^3(-4) + 6(x)^2(-4)^2 + 4(x)(-4)^3 + (-4)^4         =x4−16x3+96x2−256x+256= x^4 - 16x^3 + 96x^2 - 256x + 256

  • Exercise 1 Tasks:

    • (x+2)3(x + 2)^3

    • (x+5)4(x + 5)^4

    • (2x−1)3(2x - 1)^3

Pascal's Triangle and Coefficients

  • Definition: Pascal's triangle is a triangular array of numbers where the ends of each row are ones, and every other number is the sum of the two nearest numbers in the row above.

  • Relation to Combinations: The coefficients of binomial expansion correspond to entries in Pascal's triangle. The formula C(n,r)C(n, r) provides the value for the rr-th term in the nn-th row.

  • Examples Using Pascal's Triangle:

    • a. Evaluate C(5,2)C(5, 2): Look at row 5, term 2 of Pascal's triangle. The result is 10.

    • b. Evaluate C(6,4)C(6, 4): Look at row 6, term 4 of Pascal's triangle. The result is 15.

  • Try items:

    • Evaluate C(6,5)C(6, 5) using Pascal’s triangle.

    • Evaluate C(7,4)C(7, 4) using Pascal’s triangle.

    • Evaluate C(8,3)C(8, 3) using Pascal’s triangle.