 # Strategies for listing/calculating possible outcomes

The GCSE Maths topic of systematic listing strategies is all about working out how many possible outcomes there are – i.e. how many ways there are for something to happen – for example when finding probabilities.

Sometimes the simplest way is simply making a list of all the possible outcomes – a list like this is called the sample space – and this is normally all you need for questions at Foundation level, However, Higher questions are usually more challenging, and as the number of possibilities increases, we need to look for a different approach.

## Listing outcomes and producing sample space diagrams

If you toss a coin, what are the possible outcomes?

Now consider the possible outcomes if you toss two coins together:

This sample space diagram shows that there are 4 possible outcomes: HH, HT, TH and TT.

What would it look like for three coins? Try to work it out yourself.
Notice how, in the previous example, for each outcome on Coin 1 there are two possible outcomes on Coin 2; now extend that to include a third coin.

HHH HHT
HTH HTT
THH THT
TTH TTT
(As you can see, it’s not the most pictorial of diagrams!)
For each of the four outcomes listed in the 2-coin table, there are two possible outcomes for Coin 3 – so there are 8 possible outcomes in total.

Can you predict how many possible outcomes there would be if you were to toss six coins together?

1 coin: 2 outcomes
2 coins: 4 outcomes, and 4 = 2 × 2 = 2²
3 coins: 8 outcomes, and 8 = 2 × 2 × 2 = 2³

6 coins: the number of possible outcomes would be 2⁶, or 64.

Obviously it would take a long time to list all those!

These arrangements – where HHT, HTH and THH all count as different outcomes – are called permutations.

You don’t need to know that word for GCSE, but you do need to know that it’s important to consider whether the order matters or not! More on that later; for now we’re only looking at cases where the order DOES matter.

Now consider rolling two normal, 6-sided dice, one blue and one red. A sensible way to list the possible outcomes would be in a table like this. Finish filling it in:

How many possible outcomes are there?

Answer: 6 × 6 = 36

How many possible outcomes would there be for three dice rolled together?

Answer: 6 × 6 × 6 = 216

Now imagine tossing a coin and rolling a normal 6-sided die. How many possible outcomes are there?

Answer: 2 for the coin × 6 for the die = 12

Here’s another example: A restaurant offers a set menu with 3 choices for the starter, 4 for the main and 2 for the dessert. How many different three-course meals could you choose?

Answer: 3 × 4 × 2 = 24 different meals.

Now see if you can answer these questions.

### Exercise 1

1. If you toss a coin, roll a normal 6-sided die and draw a card at random from a standard deck of 52, how many possible outcomes are there?
2. Jamie is trying to crack a 3-digit combination lock with digits 0 to 9 in each position.
1. How many possible combinations does the lock have?
2. If he knows that the last digit is even, how many possible combinations are there?
3. If he knows that the first digit is at least 7 and the last digit is even, how many is that?
4. What if he knows that none of the digits is greater than 6?
3. A choir has 8 soprano singers, 7 altos, 6 tenors and 4 basses. One from each section is going to be chosen for a quartet performance. In how many different ways could the selection be made?
4. How many different 5-digit numbers include only the digits 6, 8 and 9?
5. How many positive integers below 1000 contain only odd digits?
6. A set menu offers 3 choices for the starter, 5 for the main course and n for the dessert. Is it possible that there are 72 different ways of choosing a starter, a main and a dessert?

In answering these questions we have been using the Product rule for counting outcomes:
The number of possible outcomes for a set of multiple events
is equal to the product of the number of possible outcomes of each event.

## Arrangements of items

Ali, Boris and Craig have three seats booked at the theatre. How many different orders can they sit in?

First, list them all:
ABC, ACB, … (you finish off the list)

That’s OK for 3 people, but would take a lot longer if it were a bigger group, so let’s consider a different approach:

In seat 1 we can have any one of the three, so there are three possible options for the occupant of seat 1.

There are only two possible occupants left for seat 2, and one for seat 3.

So the number of possible arrangements is 3 × 2 × 1 = 6.

(This can be written as “3!”, pronounced “three factorial” – but you don’t need that for GCSE – and, usefully for larger numbers, there’s an x! function on your calculator so you don’t need to type in the whole calculation.)

If there were five people and five seats then how many possible arrangements would there be?

Answer: 5 × 4 × 3 × 2 × 1 = 120.

You can see that listing all the possibilities very soon becomes impractical!

## Choosing r items from a set of n

What if we have fewer seats than people? Let’s say we have five people, A, B, C, D, E, and only three seats. How many ways are there of occupying the three seats?

Number of possible choices for seat 1 = 5
For seat 2 = 4
For seat 3 = 3

So number of possible arrangements (permutations) of 3 items from a set of 5 will be
5 × 4 × 3 = 60

## When the order doesn’t matter

If the order doesn’t matter then there will be fewer different outcomes; these are called combinations (though again, you don’t need to know that definition).

Example 1: Tossing 3 coins. Look at the list you made earlier. You have 8 possible outcomes, but some of them are duplicates if the order doesn’t matter – i.e. if you’re only interested in the total numbers of heads and tails, not which coin gave which result. How many different outcomes are there if the order is unimportant?

Answer: HHT is the same as HTH or THH, and similarly, HTT, THT and TTH are all the same, so there are only 4 possible outcomes.

Example 2: A shop sells scoops of ice cream in 8 different flavours. If I get two scoops of ice cream, both of different flavours, how many different flavour combinations could I have?

First, work out how many permutations there are (if order does matter):
8 options for scoop 1, leaving 7 options for scoop 2
So 8 × 7 = 56

Now, if you’re just having two flavours then AB is the same as BA,
and every flavour pair can be reversed in the same way,

so number of possible combinations = number of permutations ÷ 2
= 56 ÷ 2 = 28

If we also allow two scoops of the same flavour then there’s only one possible arrangement of two identical items, so in addition to the combinations of two flavours that we’ve already considered, there are 8 options where both scoops are the same (AA, BB, CC, etc.)

So in that case the total number of possible combinations is  28 + 8 = 36.

How many different flavour combinations would there be with three scoops of different flavours?

Number of permutations = 8 × 7 × 6 = 336

Number of possible arrangements of the same three flavours = 3 × 2 × 1 = 6
[It might help you to look back at Ali, Boris and Craig’s theatre trip]

So number of possible flavour combinations with 3 different scoops = 336 ÷ 6 = 56

### Exercise 2

1. Jay has 20 odd socks in his drawer. How many possible “pairs” can he make? (Left/right doesn’t matter.)
2. Eight children are playing a game where there are three prizes to be won. How many ways are there of distributing the prizes if no child wins more than one prize and
1. the prizes are all different?
2. the prizes are all identical?
3. How many different ways are there of matching 4 out of a draw’s 6 lottery numbers? (Order doesn’t matter.)
4. A class contains 16 boys and 14 girls. How many ways are there (if order doesn’t matter) of choosing:
1. two representatives, one of each gender?
2. two representatives if gender doesn’t matter?
3. three boys?
4. three students of any gender?
5. two girls and one boy?

… and that covers all you need to know about listing possible outcomes (systematic listing strategies) for Higher GCSE Maths.

For more questions on this topic, try “Product rule for counting” from MathsGenie (listed under Grade 6) and/or the “Combinations” set of exam questions collated on JustMaths.

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1) 2 × 6 × 52 = 624
2) a) 10 × 10 × 10 = 1000                b) 10 × 10 × 5 = 500
c) 3 × 10 × 5 = 150            d) 7 × 7 × 7 = 343
3) 8 × 7 × 6 × 4 = 1344
4) 3 × 3 × 3 × 3 × 3 = 243
5) 1 digit nos: 5          2 digits: 5  5 = 25           3 digits: 5 × 5 × 5 = 125            Total = 155
6) Number of possible choices = 3 × 5 × n = 15n, where n has to be an integer (whole number). 72 ÷ 15 is not an integer – so no, it is not possible.