Counting Principles & Probability — Grade 12 Maths (CAPS)

The fundamental counting principle, arrangements with n!, handling items that must stay together or that repeat, and using counting to work out a probability.

1The fundamental counting principle

If one choice can be made in m ways and the next in n ways,
the two together can be made in m × n ways.

Worked ExampleHow many 4-digit PINs are there if digits may repeat? And if they may not?

1
With repeats: 10 × 10 × 10 × 10 = 10 000.
Every position has all ten digits available.
2
Without repeats: 10 × 9 × 8 × 7 = 5 040.
Each digit used removes one option from the next position.

2Arrangements: n!

n! = n × (n − 1) × … × 2 × 1

0! = 1

n different objects can be arranged in a row in n! ways. So 5 books arrange in 5! = 120 ways, and 7 people in 7! = 5 040 ways.

3When some items must stay together

💡

Tie them into a single block, arrange everything including the block, then arrange inside the block and multiply.

Worked Example4 boys and 3 girls sit in a row. In how many ways can they sit if the girls sit together?

1
Treat the 3 girls as one block. That leaves 4 boys + 1 block = 5 items.
2
Arrange the 5 items: 5! = 120.
3
Arrange the girls inside the block: 3! = 6.
The block is not fixed internally.
4
Total = 120 × 6 = 720.

4Counting to get a probability

P(event) = (number of favourable arrangements)/(total number of arrangements)

Worked ExampleFor the 7 learners above, what is the probability that the girls sit together?

1
Favourable = 720 (from above).
2
Total arrangements of 7 people = 7! = 5 040.
No restriction.
3
P = 720/(5 040) = 1/7.
≈ 0,143.

5Repeated letters

arrangements of n letters = (n!)/(r₁! × r₂! × …)
where r₁, r₂, … are how many times each repeated letter appears

Divide by the factorial of every repeat count, not just one. If only ONE letter repeats, say r times, this is simply (n!)/(r!).

Worked ExampleHow many arrangements are there of the letters of the word LETTER?

1
There are 6 letters, but E appears twice and T appears twice.
2
(6!)/(2! × 2!) = 720/4.
Divide by the factorial of each repeat count.
3
= 180.

⚠️

Forgetting to divide by the repeats gives 720, which counts every identical-looking arrangement several times over.

6Tree diagrams still apply

A screening test: 20% of people have the disease; the test is 90% accurate on them and gives a false positive 20% of the time.

P(tests positive) = (1/5 × 9/10) + (4/5 × 2/10) = 9/50 + 8/50 = 17/50 = 0,34.

7Venn diagrams with three events

Three overlapping circles create eight regions. There is one reliable order to fill them: from the centre outwards, because every pairwise total already includes the people in the middle.

Worked ExampleOf 120 learners: 62 read novels, 55 read magazines, 48 read newspapers; 27 read novels and magazines, 22 novels and newspapers, 19 magazines and newspapers, and 11 read all three. How many read none?

1
Centre (all three) = 11
Always first.
2
Novels ∩ magazines only = 27 − 11 = 16
The 27 includes the 11 in the middle.
3
Novels ∩ newspapers only = 22 − 11 = 11; magazines ∩ newspapers only = 19 − 11 = 8
4
Novels only = 62 − 16 − 11 − 11 = 24
Subtract all three overlapping regions from the novels total.
5
Magazines only = 55 − 16 − 8 − 11 = 20; newspapers only = 48 − 11 − 8 − 11 = 18
6
At least one = 24 + 20 + 18 + 16 + 11 + 8 + 11 = 108
Add all seven inner regions.
7
None = 120 − 108 = 12

⚠️

Putting 27 straight into the novels-and-magazines overlap. That 27 is the total for both, and 11 of them also read newspapers — so only 16 belong in the two-circle region. Filling outside-in causes this every time.

8Two-way contingency tables

A contingency table splits a group by two characteristics at once. Fill in the totals row and column first — most of the missing entries then follow by subtraction.

A and B are independent ⇔ P(A and B) = P(A) × P(B)

Worked ExampleOf 200 people: 120 are male; 90 own a car, of whom 60 are male. Are 'male' and 'owns a car' independent?

1
P(male) = 120/200 = 0,6
2
P(owns a car) = 90/200 = 0,45
3
P(male AND owns a car) = 60/200 = 0,3
Read straight from the table.
4
P(male) × P(car) = 0,6 × 0,45 = 0,27
5
0,3 ≠ 0,27, so the events are NOT independent.
The test is an equality — if the two sides differ at all, the events are dependent.

💡

Independence is decided by that one equation, never by intuition about whether the two things “feel related”. Work out both sides and compare them.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
How many 4-digit PINs can be formed if digits may repeat?
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10⁴ = 10 000
2
How many 4-digit PINs can be formed if no digit may repeat?
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10 × 9 × 8 × 7 = 5 040
3
In how many ways can 5 different books be arranged on a shelf?
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5! = 120
4
4 boys and 3 girls sit in a row. In how many ways can they sit if the girls sit together?
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5! × 3! = 120 × 6 = 720
5
For those 7 learners, calculate the probability that the girls sit together.
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720/(5 040) = 1/7
6
How many arrangements are there of the letters of the word LETTER?
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(6!)/(2!×2!) = 180
7
Write down the value of 0!
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1
8
Using the tree diagram below, calculate the probability that a person tests positive.
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1/5 × 9/10 + 4/5 × 2/10 = 0,34
9
Of 80 learners: 40 play soccer, 32 play cricket, 28 play tennis; 15 play soccer and cricket, 12 soccer and tennis, 10 cricket and tennis, and 6 play all three. How many play none?
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Centre 6 ✓; pairs only 9, 6, 4 ✓; singles: soccer 40 − 9 − 6 − 6 = 19, cricket 32 − 9 − 4 − 6 = 13, tennis 28 − 6 − 4 − 6 = 12 ✓; at least one = 19 + 13 + 12 + 9 + 6 + 4 + 6 = 69; none = 80 − 69 = 11 ✓
10
In a group of 150 people, 90 are female and 60 wear glasses, of whom 40 are female. Determine whether ‘female’ and ‘wears glasses’ are independent.
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P(F) = 90/150 = 0,6 ✓; P(G) = 60/150 = 0,4 ✓; P(F and G) = 40/150 = 0,267 ✓; P(F) × P(G) = 0,24 ≠ 0,267, so NOT independent ✓
11
State the test for two events being independent.
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P(A and B) = P(A) × P(B) ✓✓
12
Why must a three-event Venn diagram be filled from the centre outwards?
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Each pairwise total already includes those in all three ✓; starting at the centre lets you subtract that overlap so nobody is counted twice ✓