Counting Methods Revision Notes for VCE SSCE Mathematical Methods

Counting Methods

Counting methods help us work out how many different ways we can arrange, select, or combine items. There are several key techniques you need to understand, each suited to different types of problems.

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The four main counting techniques are:

Addition rule - for choosing between alternatives (or)
Multiplication rule - for sequential choices (and)
Permutations - for arrangements where order matters
Combinations - for selections where order doesn't matter

The addition rule

When you need to choose between different alternatives (where you pick one option or another), you use the addition rule. Simply add together the number of choices available for each alternative.

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Key point: Use addition when the word "or" appears in the problem.

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Worked Example: Library Book Selection

Alan is at the library choosing a book to borrow. He can select from $3$ mystery novels, $3$ biographies, or $2$ science fiction books. How many total choices does he have?

Solution:

Since Alan is choosing between alternatives (mystery or biography or science fiction), we add the options:

$3 + 3 + 2 = 8$ choices

The multiplication rule

When you need to make a sequence of choices (where you pick one option and then another), use the multiplication rule. Multiply the number of options available at each stage.

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Key point: Use multiplication when the word "and" appears in the problem, indicating sequential or simultaneous choices.

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Worked Example: Outfit Combinations

Sandi is putting together an outfit. She has $6$ choices of windcheaters or jackets, and $7$ choices of jeans or skirts. How many different complete outfits can she create?

Solution:

Since Sandi needs to choose a top and a bottom, these are sequential choices, not alternatives. Using the multiplication rule:

$6 \times 7 = 42$ different outfits

We multiply because each top can be paired with each bottom, creating many combinations.

Permutations or arrangements

Permutations count the number of ways to arrange objects where order matters. For example, the arrangement ABC is different from BAC.

The number of arrangements of $n$ objects taken $r$ at a time is written as $^nP_r$ and calculated using:

^nP_r = \frac{n!}{(n-r)!} = n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)

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The factorial symbol $!$ means multiply all whole numbers from that number down to $1$ . For example, $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ .

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Worked Example: Four-Digit Numbers

How many different four-digit numbers can be formed from the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$ if each digit may be used only once?

Solution:

We're arranging $9$ digits into groups of $4$ , where order matters (since $1234$ is a different number from $4321$ ).

The number of arrangements is:

^9P_4 = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024

Therefore, 3024 different four-digit numbers can be formed.

Combinations or selections

Combinations count the number of ways to select objects where order doesn't matter. For example, selecting flavours vanilla and chocolate is the same as selecting chocolate and vanilla.

The number of combinations of $n$ objects taken $r$ at a time is written as $^nC_r$ and calculated using:

^nC_r = \frac{^nP_r}{r!} = \frac{n!}{r!(n-r)!}

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Alternative notation: You may also see this written as $\binom{n}{r}$ .

Key difference: Combinations are always smaller than or equal to permutations because we divide by $r!$ to account for the fact that order doesn't matter.

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Worked Example: Ice-Cream Selections

Four flavours of ice-cream are available at the school canteen: vanilla, chocolate, strawberry, and caramel. How many different double-scoop selections are possible if two different flavours must be used?

Solution:

We're selecting $2$ flavours from $4$ , where order doesn't matter (vanilla-chocolate is the same as chocolate-vanilla).

The number of combinations is:

^4C_2 = \frac{4!}{2! \times 2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = \frac{24}{4} = 6

Therefore, 6 different double-scoop selections are possible.

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Worked Example: Team Formation

A team of three boys and three girls must be chosen from a group of eight boys and five girls. How many different teams are possible?

Solution:

This problem requires two separate selections:

Choose $3$ boys from $8$ : $^8C_3$ ways
Choose $3$ girls from $5$ : $^5C_3$ ways

Since we need both selections and they're independent, we multiply:

^8C_3 \times ^5C_3 = 56 \times 10 = 560

Therefore, 560 different teams are possible.

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Exam Tips

Always identify whether order matters. If it does, use permutations. If it doesn't, use combinations.
Look for key words: "or" suggests addition, "and" suggests multiplication.
When calculating by hand, cancel common factors before multiplying to make calculations easier.
Double-check whether items can be repeated or must be used only once.
For team selection problems involving different groups (like boys and girls), calculate each group separately then multiply.

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Key Points to Remember:

Addition rule: Use when choosing between alternatives (or). Add the number of choices.
Multiplication rule: Use for sequential choices (and). Multiply the number of options at each stage.
Permutations ( $^nP_r$ ): Count arrangements where order matters. Formula: $\frac{n!}{(n-r)!}$
Combinations ( $^nC_r$ ): Count selections where order doesn't matter. Formula: $\frac{n!}{r!(n-r)!}$
Order matters = permutations; order doesn't matter = combinations.

Counting Methods (VCE SSCE Mathematical Methods): Revision Notes