M4.6.8 Chapter 8

Counting choices

Mei has 3 scarves and 2 hats. How many different outfits can she make with one scarf and one hat?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 3, one for each scarf.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 5, because 3 + 2.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 6, because each of the 3 scarves goes with each of the 2 hats.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Count outcomes with the product rule
Find how many ways there are to arrange 4 different letters

To find a probability by counting, you often need to know how many outcomes there are in all. Listing them works for small cases. For bigger cases, you need a faster way to count.

Start with Mei's clothes. A tree with no chances on its branches lists the outfits. Each scarf has two branches, one for each hat.

Draw it first

List the outfits as a tree

1 Draw the first choice: three branches for the three scarves.
2 From each scarf, draw two branches for the two hats. Number the ends: there are 6 outfits.

Notice that the tree has 3 groups of 2 ends. The number of ends is 3 × 2. This is the product rule. Suppose one choice can be made in a ways, and another in b ways. Then both can be made in a × b ways.

One box for each choice. Multiply the numbers in the boxes.

Worked example

Mei has 3 scarves and 2 hats. How many outfits can she make with one scarf and one hat?

There are 3 ways to choose the scarf.
For each scarf there are 2 ways to choose the hat.
Multiply: 3 × 2 = 6 outfits.

Answer: 6

The rule works for any number of choices. Multiply the number of ways for each one. A lock has 3 wheels, and each wheel shows a digit from 0 to 9. Each wheel has 10 ways.

A lock with three wheels of ten digits. 10 × 10 × 10 codes.

Worked example

A lock has 3 wheels. Each wheel shows a digit from 0 to 9. How many different codes are there?

Each wheel has 10 ways, and a wheel can repeat a digit shown by another wheel.
The first two wheels together have 10 × 10 = 100 ways.
With the third wheel, 100 × 10 = 1 000 codes.

Answer: 1 000

Worked example

Joseph guesses a code for the lock at random. What is the probability that he opens it on his first try?

All 1 000 codes are equally likely.
Only one code opens the lock.
P(opens it) = 1/1000.
As a decimal, 1 ÷ 1 000 = 0.001.

Answer: 1/1000

Now a different kind of choice: putting things in order. Each way of doing it is an arrangement. How many arrangements of the letters A, B and C in a row are there, each letter used once? A tree lists them. After you choose the first letter, only two letters are left, and after the second, only one.

Three letters in a row: 6 arrangements.

The number of choices falls each time, because a letter that is used cannot be used again. So the number of ways is 3 × 2 × 1 = 6, which agrees with the tree.

Three letters: 3 choices, then 2, then 1.

Four letters A, B, C and D work the same way. There are 4 ways to choose the first letter. Then 3 are left for the second, 2 for the third and 1 for the last.

Four different letters can be arranged in 4 × 3 × 2 × 1 = 24 ways.

Worked example

In how many ways can 4 different letters be arranged in a row?

The first place can be filled in 4 ways.
Then 3 letters are left. The first two places can be filled in 4 × 3 = 12 ways.
Then 2 letters are left, so the first three places can be filled in 12 × 2 = 24 ways.
The last place has 1 way, so 4 × 3 × 2 × 1 = 24.

Answer: 24

Counting gives probabilities. If the 4 letters are put in a random order, all 24 arrangements are equally likely. The chance that they come out as ABCD is 1 out of 24.

Try it

Find all 24 arrangements

You need: four small pieces of paper, a pencil.

Write A, B, C and D on the four pieces.
Put A first. List every way to arrange B, C and D after it. There should be 6.
Do the same with B first, then C first, then D first.
Add up. Each first letter gives 6, so the total is 4 × 6 = 24.

What you will see: Each first letter gives the same count, 3 × 2 × 1 = 6. That is the product rule at work.

If you cannot try it, this is what happens.

A wrong idea many people have

To count the outfits, add the numbers of choices: 3 + 2.

Why it feels right: We add when we put groups together, and here there are two groups, scarves and hats.

The test that breaks it: Add and you get 5, but the tree shows 6 outfits. Each scarf is worn with each hat, so every scarf is counted twice, once for each hat.

What is true: When you make one choice and then another, multiply. Add only when you choose from one group or from the other, but not both.

A wrong idea many people have

Each place in a row of 4 letters has 4 choices, so there are 4 × 4 × 4 × 4 ways.

Why it feels right: The first place does have 4 choices, and it seems the others should too.

The test that breaks it: A letter used in the first place cannot be used again. The second place has only 3 letters left, then 2, then 1.

What is true: When nothing can be used twice, each choice has one fewer option than the last: 4 × 3 × 2 × 1.

Key words

product rule: if one choice has a ways and another has b ways, both together have a × b ways.
arrangement: a way of putting things in order, such as the letters ABCD in a row.

Back to your guess

The answer is C: 6, because each of the 3 scarves goes with each of the 2 hats.

Each scarf can be worn with either hat. That makes 2 outfits for the first scarf, 2 for the second and 2 for the third. 2 + 2 + 2 = 6, which is 3 × 2.

Keep it

Copy this into your exercise book.

Product rule: a ways then b ways make a × b ways. Nothing used twice: the choices fall by one each time. 4 letters: 4 × 3 × 2 × 1 = 24.

Draw: A row of four boxes holding 4, 3, 2 and 1, with times signs between and 24 at the end.