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Multiplication

Factors and HCF Multiples and LCMDescribing probability

Independent and dependent events

This topic is relevant for:

Here we will learn about the product rule for counting, including selecting from one set and selecting from multiple sets.

There are also counting strategies worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

In combinatorics the **product rule for counting **is a method for finding the total number of ways of selecting items from a set or sets. This is part of the new GCSE specifications. Finding or listing the total number of combinations is also known as enumeration.

For example,

A teacher needs to select one boy and one girl from his class to show a visitor around the school.

There are 4 boys and 3 girls in his class.

The number of different pairs of children can be calculated by thinking about how many different options there are for each boy.

Each boy could be paired with 3 girls.

As there are 4 boys, this means there will be 4 lots of 3 pairings.

4 \times 3 = 12, so there are 12 different pairs (permutations) the teacher could select.

If the boys are called A, B, C and D and the girls are called X, Y, Z the pairings would be:

AXAYAZ

BXBYBZ

CXCYCZ

DXDYDZ

If the teacher decided to just select two girls from his class we would need to think more carefully.

There would be 3 options for the first choice, and then 2 options for the second choice. However, if calculating 3 \times 2 would give double the amount of options as it would include each pair of girls twice and the order of selecting the girls doesn’t matter.

So there are \frac{3\times 2}{2}=3 pairs of girls to select.

The possible combinations would be, XY (same as YX), XZ (same as ZX) and YZ (same as ZY).

In order to use the product rule for counting:

**Identify the number of sets to be selected from.****Identify the number of items to select from each set.****Multiply the number of items in each set. If selecting two items from a set, calculate n\times \left( n-1 \right) or \frac{n\times \left( n-1 \right)}{2} if order doesn’t matter.**

Get your free product rule for counting worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOONGet your free product rule for counting worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOON**Product rule for counting** is part of our series of lessons to support revision on **combined events**. You may find it helpful to start with the main combined events lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

Arthur has been told he can select a packet of crisps and a drink as part of a meal deal.

There are 7 different flavours of crisps and 11 different drinks.

How many possible outcomes could Arthur select?

**Identify the number of sets to be selected from.**

There are two sets to select from, crisps and drinks.

2**Identify the number of items to select from each set.**

There are 7 different crisps flavours and 11 different drinks.

3**Multiply the number of items in each set. If selecting two items from a set, calculate n\times \left( n-1 \right) or \frac{n\times \left( n-1 \right)}{2} if order doesn’t matter.**

There are 77 different pairs that Arthur could select.

Bella is clothes shopping and can select a t-shirt, a pair of jogging bottoms and hoodie as part of a mix and match offer.

There are 4 different colours of t-shirts, 5 colours for the jogging bottoms and 3 colours for the hoodie.

What is the number of outcomes Bella could select?

**Identify the number of sets to be selected from.**

There are three sets to select from.

**Identify the number of items to select from each set.**

There number of items from the three sets are 4, 5 and 3.

4\times 5\times 3=60

There are 60 different options that Bella could select.

A teacher is trying to pick two students to participate in a quiz. They must also decide which student will go first and which will go second in a round of the quiz. If there are 24 students to choose from, how many different options does the teacher have?

**Identify the number of sets to be selected from.**

There is one set to select from.

**Identify the number of items to select from each set.**

There are 24 items in the set.

Order matters because the teacher needs to select who will go first and who will go second.

24\times 23=552

There are 552 different options that the teacher could select.

15 teams in a league must play each other once. How many games in total will be played?

**Identify the number of sets to be selected from.**

There is one set to select from.

**Identify the number of items to select from each set.**

There are 15 items in the set.

Order doesn’t matter because the teams must only play each other once.

\frac{15\times 14}{2}=105

There will be 105 games played in total.

A headteacher must choose a Head Boy and Head Girl along with a Deputy Head Boy and a Deputy Head Girl.

There are 8 boys and 10 girls to select from. How many different combinations are possible?

**Identify the number of sets to be selected from.**

There are two sets to select from.

**Identify the number of items to select from each set.**

There are 8 items in the boy set and 10 in the girl set.

Order matters because there are different positions to fill.

If there are n items in one set and m items in the other set, we must calculate

n\times \left( n-1 \right)\times m\times \left( m-1 \right).

8\times 7\times 10\times 9=5040

There are 5040 different combinations.

**Forgetting to multiply by one less if selecting from a single set**

A common error is to forget that when selecting two items from a set, the second choice is from a set reduced in size by one. We must assume that the first choice is not replaced unless the question specifically states that it is.

**Not knowing when order matters or not**

Some students find it confusing to know when the order of a pair of items selected from a single set matters. If there are n items in the set and items are labelled A, B, C, …. calculating n\times \left( n-1 \right) will include the pairs AB and BA.

If the question was about teams playing each other only once, we would need divide this value by 2 to give the correct number of pairings, if, however, the teams needed to play each other twice the n\times \left( n-1 \right) would be correct.

1. There are 7 starters and 12 main courses to choose from at a restaurant. How many different ways are there of choosing a starter and main course?

77

84

72

66

For each 7 starters there are 12 possible main course choices, so calculate 7\times 12=84.

2. There are 6 starters, 8 main courses and 4 desserts to choose from at a restaurant. How many different ways are there of choosing a starter, main course and dessert?

126

80

18

192

For each 6 starters there are 8 possible main course choices and then 4 dessert choices, so calculate 6\times 8\times 4=192.

3. A teacher must choose two children from his class to show a visitor around the school. There are 22 children in his class. How many options does the teacher have to choose from?

462

484

231

242

Order doesn’t matter so calculate 22\times 21=462 and then half this answer 462\div 2=231.

4. A hockey league has 10 teams. Each team must play each other twice, home and away. How many games will be played in total?

90

100

45

50

Order matters so calculate 10\times 9=90.

5. A headteacher must choose a Head Boy and Head Girl along with a Deputy Head Boy and a Deputy Head Girl.

There are 6 boys and 7 girls to select from. How many different combinations are possible?

1764

1260

882

630

There are 6 choices for head boy and 7 for head girl. Then there are 5 choices for deputy head boy and 6 for deputy head girl.

6\times 7\times 5\times 6=1260

6. A meal deal consists of one drink and one sandwich. There are 9 drinks and x sandwiches to choose from. Which of the following **cannot** not be the amount of ways to choose a meal deal.

72

81

116

99

The amount of ways would be 9\times x, therefore it must be a multiple of 9.

1. There are 14 boys and 16 girls in a class. The teacher must choose one boy and one girl to be class representatives.

How many different ways are there of choosing one boy and one girl?

**(2 marks)**

Show answer

14\times 16

**(1)**

**(1)**

2. A combination lock has 3 dials, each with the digits 0 to 9.

Sarah chooses her combination using a set of rules.

The first number will be square.

The second number will be prime.

The third number will be odd.

(a) How many different combinations could Sarah choose?

(b) Sarah decides to include 0 as an option to each of the three numbers and says this will double the number of combinations.

Is she correct? Give a reason for your answer.

**(3 marks)**

Show answer

(a)

Sight of a list of square or prime numbers 1, 4, 9 OR 2, 3, 5, 7.

**(1)**

**(1)**

**(1)**

(b)

Yes, answer is now 4\times 5\times 6=120 .

**(1)**

3. A group of 8 people arrive for a meeting and shake each other’s hands. How many handshakes are there in total?

**(2 marks)**

Show answer

8\times 7\div 2

**(1)**

**(1)**

You have now learned how to:

- Apply use of the product rule for counting

- How to calculate probability
- Probability distribution
- Describing probability

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