In the last lesson, we learned how to organise possibilities so that nothing is missed and nothing is counted twice.
Now we are going to discover a shortcut that lets us count many possibilities without listing them all.
From a tree to multiplication
In Lesson 1, we counted outfits made from 3 T-shirts and 2 pairs of trousers.
There are 3 choices at the first stage. From each of those, there are 2 choices at the second stage.
So instead of counting the six endpoints one by one, we can write:
3 × 2 = 6
That multiplication is not a coincidence.
If a process has a choices at the first stage and b choices at the second stage, then there are a × b possible outcomes — provided every first choice can be followed by every second choice.
An ice-cream shop has 4 flavours and 3 kinds of cone.
You choose one flavour and one cone.
💡
Reveal answer
There are 4 choices of flavour. For each flavour, there are 3 choices of cone.
4 × 3 = 12
There are 12 different ice creams.
More than two stages
What if you choose two scoops of ice cream?
Suppose there are 4 flavours and repeats are allowed. For the first scoop there are 4 choices, and for the second scoop there are also 4 choices.
4 × 4 = 16
There are 16 ordered two-scoop choices. For example, chocolate then vanilla counts as different from vanilla then chocolate.
And the same idea keeps going. Suppose you are designing a character in a game. You choose:
- 3 hairstyles;
- 4 shirts;
- 2 pairs of shoes.
There are 3 choices at the first stage, 4 at the second, and 2 at the third:
3 × 4 × 2 = 24
So there are 24 different characters.
A café lets you choose:
- 2 starters;
- 5 main courses;
- 3 desserts.
You choose one of each.
💡
Reveal answer
2 × 5 × 3 = 30
There are 30 possible meals.
When listing would take far too long
Suppose a code contains 4 digits. Each digit can be any of 0, 1, 2, …, 9, and digits may repeat.
There are 10 choices for each of the four positions:
10 × 10 × 10 × 10 = 10,000
There are 10,000 possible codes — and we did not need to write down a single one.
The English alphabet has 26 letters.
A code contains three letters, and letters may repeat.
💡
Reveal answer
There are 26 choices for each position.
26 × 26 × 26 = 17,576
So there are 17,576 different three-letter codes.
What if the number of choices changes?
The multiplication idea still works when the number of choices changes from one stage to the next.
Suppose you are making a two-letter code using A, B and C, but you are not allowed to repeat a letter.
There are 3 choices for the first letter. After choosing it, only 2 letters remain.
3 × 2 = 6
The six codes are AB, AC, BA, BC, CA and CB.
A three-digit code uses the digits 1, 2, 3 and 4. No digit may be used more than once.
💡
Reveal answer
There are 4 choices for the first digit, then 3 choices, then 2.
4 × 3 × 2 = 24
There are 24 different codes.
Does the order of the stages matter?
Suppose you have 3 shirts and 4 pairs of trousers.
You could choose the shirt first:
3 × 4 = 12
Or choose the trousers first:
4 × 3 = 12
The physical order in which you make the choices does not matter. What matters is that every outfit contains one choice from each category.
A box contains some yellow balls, some blue balls and some red balls.
You make a selection by choosing one yellow ball, one blue ball and one red ball.
There are exactly 24 different ways to make such a selection.
💡
Reveal answer
We need three positive whole numbers whose product is 24.
Ignoring colour order for a moment, the possible sets are:
1, 1, 24 · 1, 2, 12 · 1, 3, 8 · 1, 4, 6 · 2, 2, 6 · 2, 3, 4
Because yellow, blue and red are different colours, swapping these numbers between the colours gives further configurations.
There are 9 balls altogether. What can you now say about the three numbers?
Reveal answer
We now need the three numbers to have product 24 and sum 9.
The only set that works is:
2, 3, 4
So there must be 2 of one colour, 3 of another, and 4 of the third.
There are more red balls than blue balls, and more blue balls than yellow balls. How many of each colour are there?
Reveal answer
2 yellow, 3 blue and 4 red.
A café offers 4 sandwiches and 3 drinks.
You may choose either one sandwich only, or one sandwich and one drink.
💡
Reveal answer
Sandwich only: 4 possibilities.
Sandwich and drink: 4 × 3 = 12 possibilities.
4 + 12 = 16
There are 16 different orders.
When one choice is followed by another, we usually multiply. When a problem splits into separate alternatives, we may need to count each case and add.
What have we learned?
When choices happen in stages, we can often count the possibilities by multiplying the number of choices at each stage.
a × b × c × ···
But always ask:
- How many choices are really available at each stage?
- Do earlier choices remove later possibilities?
- Does the problem split into different cases?
Next: Arrangements
What happens when we arrange several different objects in order?
Why do the numbers 1, 2, 6, 24, 120, … keep appearing?
That leads us to factorials and permutations.
What did you think?
Whether you explored this together or worked independently, we would love to hear from both parents and young learners. Tell us what you enjoyed, what was confusing, or what you would like to see next.
Share your feedback
The form is short and you do not need to give your name.