FAD1015 Week 1 — Counting Rules & Permutation

Week 1 lecture covering fundamental counting principles, permutations, combinations, and basic probability. The lecture is divided into two parts: LEC 1 focuses on counting rules, while LEC 2 introduces probability concepts.

Summary

• Multiplication Rule for sequential choices across multiple categories
• Permutations — ordered arrangements of distinct objects, with repetition, with identical objects, with restrictions, and in a circle
• Combinations — unordered selections where order does not matter
• Theoretical Probability — $P(E) = n(E)/n(S)$, sample spaces, events, and outcomes
• Compound Events — intersection ($A \cap B$) and union ($A \cup B$), addition rule

LEC 1: Counting Rules

1. Multiplication Rules

A simple method of finding the number of ways using the multiplication rule.

If there are $n, m, p, r, s$ ways to choose distinct objects of category 1, 2, 3, 4 and 5 respectively, then the number of ways to choose one object from each category is: $$n \times m \times p \times r \times s$$

Example 1: How many ways are there to choose a book and a magazine from 6 different books and 7 different magazines?

$$6 \times 7 = 42 \text{ ways}$$

Example 2: How many ways can a customer choose a meal comprising rice, a vegetable, a main dish and a dessert from a choice of 2 types of rice, 5 kinds of vegetable, 8 different main dishes and 3 kinds of dessert?

$$2 \times 5 \times 8 \times 3 = 240 \text{ ways}$$

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2. Permutation

Permutation is the arrangement of some or all distinct objects that considers the order of the objects.

For example, the arrangement $AB$ is different from $BA$ (i.e. $BA \neq AB$).

2.1 Permutations of $n$ Distinct Objects

The number of permutations of arranging $n$ distinct objects is given by:

$$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$$

Note: $!$ is called the factorial operator and $\boxed{0! = 1}$.

Example: The permutations of two digits from $A, B, C$ are: $$AB, BA, BC, CB, AC, CA \quad \Rightarrow \quad 3! = 6 \text{ ways}$$

2.2 Permutations of $r$ from $n$ Distinct Objects

The number of permutations of arranging $r$ out of $n$ distinct objects is given by:

$$P(n,r) = {}_n P_r = \frac{n!}{(n-r)!}$$

Example 5 (from lecture): [Standard arrangement problem using $P(n,r)$]

Exercise 3: A four-digit PIN is selected. How many different PINs are possible if:

• (i) repetition is allowed? $\quad 10^4 = 10,000$
• (ii) repetition is not allowed? $\quad P(10,4) = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040$

2.3 Permutations with Repetition Allowed

When repetitions are allowed, the number of arrangements of $r$ objects chosen from $n$ distinct objects is:

$$n^r$$

2.4 Permutations with Identical Objects

When arranging $n$ objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type $k$:

$$\frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}$$

Example 6: How many ways are there to arrange all letters from the word:

• (a) PROBABILITY — $\frac{11!}{2! \times 2!}$ (two B's, two I's)
• (b) STATISTICS — $\frac{10!}{3! \times 3! \times 2!}$ (three S's, three T's, two I's)

2.5 Arrangements with Restrictions

Example 7: Naura has 11 different CDs: 6 pop, 3 jazz, 2 classical. How many arrangements of all 11 CDs on a shelf are there if the jazz CDs are all next to each other?

Treat the 3 jazz CDs as one block/unit. Then we arrange: 6 pop + 1 jazz block + 2 classical = 9 units $$9! \times 3!$$ (The $3!$ accounts for arrangements within the jazz block.)

Example 8: Four-digit numbers are formed from ${1, 2, 3, 4, 5, 6}$. How many arrangements:

• (a) are greater than 2000? $\quad 5 \times 6 \times 6 \times 6 = 1080$ (first digit can be 2–6)
• (b) are even? $\quad 6 \times 6 \times 6 \times 3 = 648$ (last digit must be 2, 4, or 6)
• (c) do not start with 5? $\quad 5 \times 6 \times 6 \times 6 = 1080$

Example 9: Five girls and seven boys sit in a row. How many arrangements:

• (a) no specific preference? $\quad 12!$
• (b) all girls must sit next to each other? $\quad 8! \times 5!$ (treat 5 girls as one block, giving 8 units)

Example 10: Eight members stand in a line, but Nayla and Dura refuse to stand next to each other.

Total arrangements: $8!$ Arrangements where Nayla and Dura ARE together: $7! \times 2!$

$$\text{Valid arrangements} = 8! - 7! \times 2! = 40,320 - 10,080 = 30,240$$

2.6 Permutation in a Circle

For circular arrangements, rotations of the same arrangement are considered identical.

The number of permutations of arranging $n$ different objects in a circle:

• Where clockwise and anticlockwise are considered different: $(n-1)!$
• Where clockwise and anticlockwise are considered the same (e.g. a ring, necklace): $\frac{(n-1)!}{2}$

Example 11: In how many ways can a group of 6 people sit at a round table? $$(6-1)! = 5! = 120$$

Example 12: In how many ways can 10 different coloured beads be arranged to form a ring? $$\frac{(10-1)!}{2} = \frac{9!}{2} = 181,440$$ (Flip the ring and it's the same arrangement.)

Example 13: How many arrangements of the letters from SCIENCE around a circle?

SCIENCE has 7 letters with two C's identical. Linear arrangements: $\frac{7!}{2!}$ Circular arrangements (clockwise/anticlockwise different): $\frac{7!}{2! \times 7} = \frac{6!}{2!} = 360$

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3. Combination

Combination is the selection of objects where order does not matter.

The number of combinations of choosing $r$ objects from $n$ distinct objects is:

$$C(n,r) = {}_n C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

Example 15: A team of two is chosen from 1 male and 2 female students. In how many ways?

$$C(3,2) = \frac{3!}{2!1!} = 3 \text{ ways}$$

Example 16: A team of three is chosen from 6 male and 8 female students. How many ways if there must be more female than male? Hence find the probability.

Cases with more female than male:

• 0 male, 3 female: $C(6,0) \times C(8,3) = 1 \times 56 = 56$
• 1 male, 2 female: $C(6,1) \times C(8,2) = 6 \times 28 = 168$
• 2 male, 1 female: NOT valid (not more female)

Total valid: $56 + 168 = 224$ Total possible teams: $C(14,3) = 364$

$$P(\text{more female than male}) = \frac{224}{364} = \frac{8}{13}$$

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4. Permutation vs Combination

Permutation	Combination
Order matters	Order does not matter
Arrangement of objects	Selection of objects
$P(n,r) = \frac{n!}{(n-r)!}$	$C(n,r) = \frac{n!}{r!(n-r)!}$
$AB \neq BA$	$AB = BA$
PIN codes, race positions, seating arrangements	Committees, teams, lottery numbers

graph TD
    Q1["Does order matter?"] -->|Yes| Q2["Are all objects used?"]
    Q1 -->|No| COMB["Combination<br/>C(n,r) = n! / (r!(n-r)!)"]
    Q2 -->|Yes| PERM["Permutation of n distinct objects<br/>n!"]
    Q2 -->|No, only r objects| Q3["Are repetitions allowed?"]
    Q3 -->|Yes| REP["Permutation with repetition<br/>n^r"]
    Q3 -->|No| Q4["Are there identical objects?"]
    Q4 -->|Yes| IDENT["Permutation with identical objects<br/>n! / (n_1! n_2! ... n_k!)"]
    Q4 -->|No| NPR["Permutation P(n,r)<br/>n! / (n-r)!"]
    Q4 -->|Circular| CIRC["Circular permutation<br/>(n-1)! or (n-1)!/2"]

    style Q1 fill:#e7f5ff,stroke:#1971c2
    style COMB fill:#d3f9d8,stroke:#2f9e44
    style PERM fill:#ffe8cc,stroke:#d9480f
    style REP fill:#ffe8cc,stroke:#d9480f
    style IDENT fill:#ffe8cc,stroke:#d9480f
    style NPR fill:#ffe8cc,stroke:#d9480f
    style CIRC fill:#ffe8cc,stroke:#d9480f

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LEC 2: Probability

graph LR
    subgraph exp["Probability Experiment"]
        E1["Toss a coin"]
        E2["Roll a die"]
        E3["Draw a card"]
    end
    subgraph outcomes["Outcomes"]
        O1["H, T"]
        O2["1, 2, 3, 4, 5, 6"]
        O3["52 cards"]
    end
    subgraph sample["Sample Space S"]
        S1["{H, T}"]
        S2["{1, 2, 3, 4, 5, 6}"]
        S3["{all 52 cards}"]
    end
    subgraph event["Event E"]
        EV1["Getting a Head"]
        EV2["Rolling an even number"]
        EV3["Drawing a Spade"]
    end

    E1 --> O1 --> S1 --> EV1
    E2 --> O2 --> S2 --> EV2
    E3 --> O3 --> S3 --> EV3

1. Experiment, Sample Space, Event and Outcome

• Probability experiment: A process that involves chance which gives several possible results known as outcomes.
• Outcome: A result of a trial of an experiment.
• Sample space ($S$): The set of all possible outcomes of a probability experiment.
• Event: A set consisting of one or more outcomes of a probability experiment.

Example 17: Find the sample space of:

• (a) Two coins are tossed: $S = {HH, HT, TH, TT}$
• (b) Tossing a coin and rolling a die: $S = {(H,1), (H,2), \dots, (H,6), (T,1), \dots, (T,6)}$
• (c) Taking a driving test: $S = {\text{Pass}, \text{Fail}}$

Example 18: Two fair dice are tossed. List outcomes of:

• (a) Both even numbers: ${(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}$
• (b) Both the same number: ${(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}$

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2. Theoretical Probability

For any event $E$ in sample space $S$:

$$P(E) = \frac{n(E)}{n(S)} = \frac{\text{number of outcomes resulting in event } E}{\text{number of outcomes in sample space } S}$$

Properties:

• A probability of $0$ indicates the event is impossible.
• A probability of $1$ (or 100%) indicates the event is certain.
• All other events have probability between $0$ and $1$.

Example 19: A box contains 20 cards numbered 1 to 20. A card is picked at random.

• (a) $P(\text{multiple of 5}) = \frac{4}{20} = \frac{1}{5}$
• (b) $P(\text{not a multiple of 5}) = 1 - \frac{1}{5} = \frac{4}{5}$
• (c) $P(\text{higher than 7}) = \frac{13}{20}$

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3. Playing Cards

An ordinary pack consists of 52 cards, split equally into four suits:

Colour	Suits
Red	Diamonds ($\diamondsuit$), Hearts ($\heartsuit$)
Black	Clubs ($\clubsuit$), Spades ($\spadesuit$)

Each suit has 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

The Jack, Queen and King of any suit are called picture cards.

Example 20: A card is dealt from a well-shuffled pack of 52.

• (a)(i) $P(\text{5 of clubs}) = \frac{1}{52}$
• (a)(ii) $P(\text{5 of clubs or any diamond}) = \frac{1}{52} + \frac{13}{52} = \frac{14}{52} = \frac{7}{26}$
• (b) First card is 7 of spades. $P(\text{second card is black suit}) = \frac{25}{51}$

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4. Two-way Table

Example 21:

	Male	Female	Total
Pass	57	62	119
Fail	12	19	31
Total	69	81	150

• (a) $P(\text{failed}) = \frac{31}{150}$
• (b) $P(\text{female and passed}) = \frac{62}{150} = \frac{31}{75}$
• (c) $P(\text{failed } | \text{ male}) = \frac{12}{69} = \frac{4}{23}$

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5. Probability Using Counting Methods

Example 22: The letters of MATRICULATION are arranged at random in a line.

• (a) Total arrangements: $\frac{13!}{2!}$ (two I's are identical)
• (b)(i) $P(\text{both T's are together}) = \frac{12!}{2!} \div \frac{13!}{2!} = \frac{12!}{13!} = \frac{1}{13}$
• (b)(ii) $P(\text{both T's are separated}) = 1 - \frac{1}{13} = \frac{12}{13}$
• (c) $P(\text{all vowels together})$: treat 6 vowels (A, I, I, A, I, O) as one block. 8 units total. Vowels within block: $\frac{6!}{3!2!}$. Total valid: $8! \times \frac{6!}{3!2!}$

Example 23: A committee of 5 is chosen from 6 boys and 4 girls.

• (a) $P(\text{3 boys and 2 girls}) = \frac{C(6,3) \times C(4,2)}{C(10,5)}$
• (b) $P(\text{more boys than girls}) = \frac{C(6,4)C(4,1) + C(6,5)}{C(10,5)}$
• (c) $P(\text{3 boys and 2 girls, but girl X refuses to serve with boy Y})$: Total valid committees minus those containing both X and Y.

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6. Probability of Compound Events

For two events $A$ and $B$ in the sample space:

• Intersection ($A \cap B$): both $A$ and $B$ occur $$P(A \text{ and } B) = P(A \cap B)$$

• Union ($A \cup B$): at least one of $A$ or $B$ occurs $$P(A \text{ or } B) = P(A \cup B)$$

Addition Rule for Combined Events

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

If $A$ and $B$ are mutually exclusive (cannot occur together), then $P(A \cap B) = 0$ and:

$$P(A \cup B) = P(A) + P(B)$$

Example 24: A number is selected from the first 250 positive integers. What is the probability that it is exactly divisible by 2 or 7?

$$P(2 \text{ or } 7) = P(2) + P(7) - P(14) = \frac{125}{250} + \frac{35}{250} - \frac{17}{250} = \frac{143}{250}$$

Example 25: A fair die is rolled once. Find the probability of getting:

• (a) an even number: $\frac{3}{6} = \frac{1}{2}$
• (b) a number less than 5: $\frac{4}{6} = \frac{2}{3}$
• (c) divisible by 2: $\frac{3}{6} = \frac{1}{2}$
• (d) not divisible by 2: $1 - \frac{1}{2} = \frac{1}{2}$
• (e) an even number or a number less than 5: $$P(\text{even} \cup <5) = P(\text{even}) + P(<5) - P(\text{even and } <5) = \frac{3}{6} + \frac{4}{6} - \frac{2}{6} = \frac{5}{6}$$ (The overlap is ${2, 4}$.)

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Related Topics

• FAD1015 Week 2 — Mutually Exclusive & Conditional Probability — conditional probability and independence
• FAD1015 Week 3 — Independent Events & Bayes' Theorem — Bayes' theorem and total probability
• FAD1015 Tutorial 1-6 — Counting & Probability Fundamentals — practice problems

Related Course Page

• FAD1015 - Mathematics III