Listing Outcomes Systematically

To find a probability you often need to know how many outcomes there are. Listing them at random almost guarantees you will miss some or repeat others.

The golden rule

Fix an order and stick to it — change the last item first, then work backwards

Worked Example 1 — Listing in order

A café offers two sizes (Small, Large) and three flavours (Chocolate, Vanilla, Strawberry). List all the possible orders.

①Fix the first item as Small and run through every flavour.

②SC, SV, SS

③Now fix the first item as Large and repeat.

④LC, LV, LS

Six possible orders altogether.

Notice the structure: the first letter stays fixed while the second cycles through all its options. That pattern makes it obvious when you have finished.

Product rule

If there are $m$ ways of doing the first thing and $n$ ways of doing the second,
there are $m \times n$ ways of doing both

This extends to any number of stages: multiply the number of choices at each stage together.

Worked Example 2 — Counting without listing

A menu has $4$ starters, $6$ mains and $3$ desserts. How many different three-course meals are possible?

①$4$ choices, then $6$, then $3$.

②$4 \times 6 \times 3 = 72$ meals

Listing all $72$ would be tedious; the product rule gives the count instantly.

Worked Example 3 — Using the count in a probability

A four-digit PIN uses digits $0$–$9$, and digits may repeat. Find (a) how many PINs are possible, (b) the probability of guessing one correctly at random.

①(a) Each of the four positions has $10$ choices.

②$10 \times 10 \times 10 \times 10 = 10\,000$ PINs

③(b) $P(\text{correct guess}) = \dfrac{1}{10\,000} = 0.0001$

Worked Example 4 — When repeats are not allowed

Three students are chosen from a group of $8$ to be first, second and third in a queue. How many orders are possible?

①First place: $8$ choices.

②Second place: only $7$ left.

③Third place: only $6$ left.

④$8 \times 7 \times 6 = 336$ orders

Read whether repeats are allowed. "Digits may repeat" gives $10 \times 10$; "all digits different" gives $10 \times 9$. This single word changes the answer completely.

A Venn diagram shows how sets overlap. It is the clearest way to handle "both", "either" and "neither" questions.

Always fill in the overlap first — the "both" region.
Subtract to find "only A" and "only B".
Add the three regions and subtract from the total to find "neither".
Check that all four regions add to the grand total.

Start with the middle. If $18$ study French and $7$ study both, then "French only" is $18 - 7 = 11$, not $18$. The $18$ already includes the seven who do both.

Worked Example 5 — Filling in a Venn diagram

In a group of $50$ people, $28$ own a bike, $19$ own a scooter and $9$ own both. Complete the Venn diagram and find $P(\text{owns neither})$.

①Overlap (both) $= 9$.

②Bike only $= 28 - 9 = 19$.

③Scooter only $= 19 - 9 = 10$.

④At least one $= 19 + 9 + 10 = 38$.

⑤Neither $= 50 - 38 = 12$.

$P(\text{neither}) = \dfrac{12}{50} = 0.24$

Worked Example 6 — Working backwards

Of $40$ students, $25$ play football, $6$ play neither football nor tennis, and $14$ play tennis. How many play both?

①At least one sport $= 40 - 6 = 34$.

②If nobody played both, the total would be $25 + 14 = 39$.

③But only $34$ play at least one, so the excess $39 - 34 = 5$ have been counted twice.

④$5$ students play both.

Check: football only $= 20$, both $= 5$, tennis only $= 9$, neither $= 6$; total $= 40$ ✓

Symbol	Name	Meaning
$\xi$	Universal set	Everything being considered
$A \cap B$	Intersection	In $A$ and $B$ — the overlap
$A \cup B$	Union	In $A$ or $B$ (or both)
$A'$	Complement	Not in $A$
$n(A)$	Number of elements	How many things are in $A$
$\emptyset$	Empty set	Contains nothing

Remembering the symbols: $\cap$ looks like a cap or a bridge — you need to be under both ends, so it means AND. $\cup$ is a cup that collects everything, so it means OR.

Worked Example 7 — Reading set notation

Using the data from Worked Example 5 ($50$ people; bike only $19$, both $9$, scooter only $10$, neither $12$), find:

(a) $n(B \cap S)$ (b) $n(B \cup S)$ (c) $n(B')$ (d) $P(B \cap S')$

①(a) The intersection is "both": $9$.

②(b) The union is everyone in at least one set: $19 + 9 + 10 = 38$.

③(c) Not a bike owner: $50 - 28 = 22$ (scooter only plus neither $= 10 + 12$ ✓).

④(d) Bike but not scooter — that is "bike only", $19$.

⑤$P = \dfrac{19}{50} = 0.38$

The order to fill it in

Centre (all three) → the three pairwise overlaps → the three "only" regions → outside

Worked Example 8 — Three sets

$60$ students were asked which of Maths, Physics and Chemistry they study. $8$ study all three. $15$ study Maths and Physics, $12$ study Physics and Chemistry, $11$ study Maths and Chemistry. Altogether $34$ study Maths, $30$ study Physics and $25$ study Chemistry. How many study none of them?

①Centre: all three $= 8$.

②Maths and Physics only $= 15 - 8 = 7$.

③Physics and Chemistry only $= 12 - 8 = 4$.

④Maths and Chemistry only $= 11 - 8 = 3$.

⑤Maths only $= 34 - 7 - 8 - 3 = 16$.

⑥Physics only $= 30 - 7 - 8 - 4 = 11$.

⑦Chemistry only $= 25 - 3 - 8 - 4 = 10$.

⑧Total in at least one $= 16 + 7 + 11 + 3 + 8 + 4 + 10 = 59$.

⑨None $= 60 - 59 = \mathbf{1}$ student.

Be systematic

Fix an order; change the last item first.

Product rule

$m$ ways then $n$ ways gives $m \times n$ altogether.

Repeats?

Allowed: $10 \times 10$. Not allowed: $10 \times 9$.

Venn: start in the middle

Fill the overlap first, then subtract outwards.

$\cap$ intersection

AND — the overlap.

$\cup$ union

OR — everything in either set.

$A'$

Not $A$ — everything outside the circle.

Three sets

Centre, then pairs, then singles, then outside.

Always check

All regions must add to the grand total.

Question 1

A shop sells jumpers in $3$ colours and $4$ sizes. How many different jumpers are there?

▶ Show solution

$3 \times 4 = 12$ different jumpers

Question 2

List systematically all the two-digit numbers that can be made using the digits $2$, $5$ and $7$, if digits may be repeated. How many are there?

▶ Show solution

Fix the first digit and cycle the second:

$22, 25, 27$; $52, 55, 57$; $72, 75, 77$

That is $3 \times 3 = 9$ numbers.

Question 3

Repeat Question 2 but with all digits different. How many numbers now?

▶ Show solution

$25, 27, 52, 57, 72, 75$

$3 \times 2 = 6$ numbers.

Question 4

A password has $3$ letters (A–Z) followed by $2$ digits (0–9), with repeats allowed. How many passwords are possible?

▶ Show solution

$26 \times 26 \times 26 \times 10 \times 10$

$= 17\,576 \times 100 = 1\,757\,600$ passwords

Question 5

In a group of $40$ people, $22$ like tea, $17$ like coffee and $8$ like both. Find how many like neither.

▶ Show solution

Tea only $= 22 - 8 = 14$; coffee only $= 17 - 8 = 9$.

At least one $= 14 + 8 + 9 = 31$.

Neither $= 40 - 31 = 9$ people.

Question 6

Using the data from Question 5, find (a) $P(\text{likes tea only})$, (b) $n(T \cup C)$, (c) $n(C')$.

▶ Show solution

(a) $\dfrac{14}{40} = 0.35$

(b) $31$ — everyone who likes at least one drink.

(c) Not coffee $= 40 - 17 = 23$ (tea only $14$ plus neither $9$ ✓).

Question 7

Of $70$ people surveyed, $45$ own a car, $12$ own neither a car nor a bike, and $30$ own a bike. How many own both?

▶ Show solution

At least one $= 70 - 12 = 58$.

If nobody owned both: $45 + 30 = 75$.

Double-counted $= 75 - 58 = 17$.

$17$ people own both.

Check: car only $28$, both $17$, bike only $13$, neither $12$; total $= 70$ ✓

Question 8

Four athletes run a race. In how many different orders can they finish, assuming no ties?

▶ Show solution

First place: $4$ choices; second: $3$; third: $2$; fourth: $1$.

$4 \times 3 \times 2 \times 1 = 24$ orders

Question 9

A restaurant offers $5$ starters, $8$ mains and $4$ desserts.

(a) How many three-course meals are possible? (b) How many meals are possible if you may skip the starter? (c) A customer picks at random. Find the probability they choose one particular three-course combination.

▶ Show solution

(a) $5 \times 8 \times 4 = 160$ meals

(b) Skipping the starter is now a sixth option for that course: $6 \times 8 \times 4 = 192$ meals.

(c) $\dfrac{1}{160} = 0.00625$

Question 10

$100$ students were asked about three clubs: Drama, Sport and Music. $12$ do all three; $25$ do Drama and Sport; $20$ do Sport and Music; $18$ do Drama and Music. In total $50$ do Drama, $55$ do Sport and $40$ do Music.

(a) How many do Sport only? (b) How many do none of the three? (c) A student is chosen at random; find $P(\text{exactly two clubs})$.

▶ Show solution

Fill in from the centre:

All three $= 12$.

Drama and Sport only $= 25 - 12 = 13$.

Sport and Music only $= 20 - 12 = 8$.

Drama and Music only $= 18 - 12 = 6$.

(a) Sport only $= 55 - 13 - 12 - 8 = 22$

(b) Drama only $= 50 - 13 - 12 - 6 = 19$; Music only $= 40 - 6 - 12 - 8 = 14$.

At least one $= 19 + 13 + 22 + 6 + 12 + 8 + 14 = 94$.

None $= 100 - 94 = \mathbf{6}$ students.

(c) Exactly two means the three pairwise-only regions: $13 + 8 + 6 = 27$.

$P = \dfrac{27}{100} = 0.27$