EDEXCEL GCSE MATHS · FOUNDATION & HIGHER
Number problems and reasoning
Number · Lesson 1 of 7
Before We Start
Answer each one, then check.
1. Work out 3 × 4 × 5.
60
2. Work out 6 × 5 × 4.
120
3. A coin is flipped. How many possible outcomes are there?
2: heads or tails.
4. A dice is rolled. How many possible outcomes are there?
6
5. List every two-digit number you can make from the digits 1 and 2, using each digit once.
12 and 21
Learning Objectives
1. List all the possible outcomes of a situation systematically.
2. Use the product rule to count outcomes without listing them. (Higher)
3. Count arrangements when items cannot be used twice. (Higher)
4. Solve counting problems with restrictions. (Higher)
5. Count the pairs that can be chosen from a group. (Higher)
Listing Systematically
A systematic list follows a pattern, so that no outcome is missed and none is written twice.
▸ Fix the first choice. Keep the first item the same and work through every option for the second.
▸ Then move on. Change the first item and repeat the same pattern.
▸ Use letters. Write S for soup, C for chicken and so on - it is quicker and clearer.
▸ Count at the end. Once the list is complete, count it.
A Systematic List
A café offers 2 starters (Soup, Melon) and 3 mains (Chicken, Fish, Pasta).
|
Starter |
With Chicken |
With Fish |
With Pasta |
|---|---|---|---|
|
Soup (S) |
SC |
SF |
SP |
|
Melon (M) |
MC |
MF |
MP |
|
Total |
2 rows |
3 columns |
2 × 3 = 6 meals |
PART ONE · HIGHER
The Product Rule
Counting without listing.
The Product Rule for Counting
If one choice can be made in m ways and a second choice in n ways, the two choices together can be made in m × n ways.
▸ Two choices. 2 starters and 3 mains give 2 × 3 = 6 meals - exactly the list we wrote.
▸ More choices. Keep multiplying: 5 sandwiches, 4 snacks and 3 drinks give 5 × 4 × 3 = 60 meal deals.
▸ Why it works. Every one of the first choices can be paired with every one of the second.
▸ When to use it. When the choices are made one after another and each choice does not affect the others.
A Tree Diagram of Choices
|
A tree diagram shows why the product rule works. Each of the 2 first branches splits into 3, so there are 2 × 3 = 6 routes from start to finish - one for every possible meal. |
Each starter branches into 3 mains: 2 × 3 = 6 outcomes. |
Counting Outfits
|
Priya has 4 T-shirts, 3 pairs of jeans and 2 pairs of trainers. How many different outfits of one T-shirt, one pair of jeans and one pair of trainers can she make? |
1. Count the choices for the T-shirt
4
2. Count the choices for the jeans
3
3. Count the choices for the trainers
2
4. The choices are independent, so multiply
4 × 3 × 2 = 24
Answer: 24 different outfits
Counting Codes
|
A padlock code uses 4 digits from 0 to 9. (a) How many codes are possible? (b) How many are possible if no digit can be used twice? |
1. (a) Each position can be any of 10 digits
10 × 10 × 10 × 10
2. Work it out
= 10 000
3. (b) The first digit has 10 choices
10
4. Each digit used leaves one fewer for the next
10 × 9 × 8 × 7
5. Work it out
= 5040
Answer: (a) 10 000 codes (b) 5040 codes
PART TWO · HIGHER
Arrangements and Restrictions
When items cannot be used twice, and some positions have rules.
Arranging Items in a Row
When items cannot be repeated, the number of choices goes down by one each time.
▸ 3 people in a line. 3 × 2 × 1 = 6 ways.
▸ 5 people in a line. 5 × 4 × 3 × 2 × 1 = 120 ways.
▸ The pattern. n different items can be arranged in n × (n−1) × … × 2 × 1 ways.
▸ Your calculator. The x! key (factorial) works this out: 5! = 120.
Counting with a Restriction
|
How many three-digit numbers can be made from the digits 2, 3, 5 and 8, using each digit at most once, if the number must be even? |
1. Deal with the restricted position first: the last digit must be even
2 or 8: 2 choices
2. The first digit can be any of the 3 digits left
3 choices
3. The middle digit can be any of the 2 digits left
2 choices
4. Multiply the choices
2 × 3 × 2 = 12
Answer: 12 even three-digit numbers
Choosing Pairs
When two people are chosen and the order does not matter, each pair gets counted twice.
▸ Handshakes. 10 people each shake hands with everyone else once. 10 × 9 = 90 counts every handshake twice (A with B, and B with A), so there are 90 ÷ 2 = 45 handshakes.
▸ The pattern. n people make n(n−1)/2 pairs.
▸ When order matters. Choosing a captain and a vice-captain from 10 people: 10 × 9 = 90 ways. Here A-then-B is different from B-then-A, so do not halve.
Does the Order Matter?
|
ORDER MATTERS - DO NOT HALVE |
ORDER DOES NOT MATTER - HALVE FOR PAIRS |
|
▸ Captain and vice-captain. ▸ 1st and 2nd place in a race. ▸ Codes and PINs: 12 is not 21. ▸ Arranging people in a queue. |
▸ Handshakes between two people. ▸ Choosing two people for a team. ▸ Two games in a round-robin chess club. ▸ Picking 2 toppings from a list. |
Counting Problem Checklist
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Are repeats allowed? Codes usually allow repeats; people in a queue cannot be used twice. |
Does the order matter? If swapping two choices gives the same outcome, you have double-counted. |
|
Is there a restriction? Fill the restricted positions first, then the rest. |
Check with a small case Try the method on a small version you can list, and see if it gives the same count. |
Case Study
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CASE STUDY The Rubik's Cube The Hungarian architecture lecturer Ernő Rubik invented his cube in 1974. It has 26 small visible pieces, and the product rule shows how the number of arrangements explodes: the 8 corner pieces alone can be placed in 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40 320 orders, before they are even twisted. Multiply in the ways of placing and twisting the corners and edges, allow for the positions that cannot actually be reached, and the total number of positions is about 43 quintillion - 43 252 003 274 489 856 000. |
|
1974 Ernő Rubik invents the cube |
4.3 × 10¹⁹ Possible positions of a standard 3 × 3 × 3 cube |
Key Terms
|
Outcome One possible result of a choice or experiment. |
Systematic list A list that follows a pattern so no outcome is missed or repeated. |
|
Product rule for counting If there are m ways to do one thing and n ways to do another, there are m × n ways to do both. (Higher) |
Arrangement An ordering of items, for example people in a queue. |
|
Restriction A rule that limits the choices, such as "the number must be even". |
Factorial n! = n × (n−1) × … × 2 × 1. For example, 4! = 24. |
Your Task: The Ice Cream Van
10 minutes
|
An ice cream van sells 6 flavours, 3 types of cone and 4 toppings. Work out: (a) how many ice creams with one flavour, one cone and one topping are possible; (b) how many there are if the customer chooses two different flavours, and the order of the flavours does not matter; (c) how many two-flavour ices there are if the order of the scoops DOES matter. 1. Part (a): one flavour. 2. Part (b): two flavours, order does not matter. 3. Part (c): two flavours, order matters. |
A good answer shows: (a) 6 × 3 × 4 = 72. (b) Pairs of flavours: 6 × 5 ÷ 2 = 15, so 15 × 3 × 4 = 180. (c) 6 × 5 = 30, so 30 × 3 × 4 = 360.
Can I...?
☐ List outcomes systematically.
☐ Use a table to list combinations.
☐ Use the product rule for counting. (Higher)
☐ Count codes with and without repeats. (Higher)
☐ Count arrangements of items in a row. (Higher)
☐ Fill a restricted position first. (Higher)
☐ Count pairs chosen from a group. (Higher)
☐ Decide whether the order matters. (Higher)
Summary
✓ List systematically: fix the first choice and work through the rest.
✓ Product rule (Higher): m ways and n ways give m × n ways altogether.
✓ No repeats: the choices go down by one each time.
✓ Restrictions: fill the restricted position first.
✓ Pairs where order does not matter: n(n−1)/2.
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EXAM FOCUS There are 12 students in a chess club. Each student plays every other student once. How many games are played? (3 marks) Write the calculation as well as the answer: the method marks are for the product (12 × 11) and for dividing by 2 because each game was counted twice. |