By now you have several different counting tools: lists, tables, trees, multiplication, factorials, combinations and Pascal’s triangle.
The final skill is learning when not to count a problem in the obvious way.
Count what is easier
Suppose a four-digit code can use any digit from 0 to 9, and repetition is allowed.
How many codes contain at least one 7?
Trying to count codes with exactly one 7, then exactly two 7s, and so on would work — but there is a much easier route.
There are
104 = 10,000
four-digit codes altogether.
If a code contains no 7 at all, each digit has only 9 choices:
94 = 6,561
So the number containing at least one 7 is
10,000 − 6,561 = 3,439
If every possibility is either wanted or unwanted, then:
wanted = total − unwanted
A coin is tossed 5 times.
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Reveal answer
There are 25 = 32 sequences altogether.
Only one contains no heads: TTTTT.
32 − 1 = 31
So there are 31 sequences with at least one head.
Split into cases
Another useful strategy is to split a problem into separate cases that cannot overlap.
Suppose a café offers 4 sandwiches and 3 salads. You may order either one sandwich or one salad.
There are 4 sandwich orders and 3 salad orders, so:
4 + 3 = 7
When one choice is followed by another, multiply. When a problem divides into distinct alternatives, count each alternative and add.
A two-symbol code is made in one of two ways:
- a letter followed by a digit;
- or a digit followed by a letter.
There are 26 letters and 10 digits.
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Reveal answer
Letter then digit: 26 × 10 = 260.
Digit then letter: 10 × 26 = 260.
260 + 260 = 520
There are 520 codes.
Use symmetry
Sometimes several parts of a problem must have the same answer by symmetry.
Imagine choosing 3 people from 8. The number of groups containing Alice is the same as the number containing Ben, Charlie or any other particular person.
Recognising symmetry can save a lot of repeated work.
All 3-person teams are formed from 6 children.
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Reveal answer
No. By symmetry, every child must appear in the same number of teams.
In fact, fixing one child leaves 2 teammates to choose from the remaining 5, so each child appears in 10 teams.
Count the same thing in two ways
One of the most powerful counting tricks is to count the same collection in two different ways.
Suppose 8 people each shake hands with every other person exactly once.
From each person’s point of view there are 7 handshakes, so:
8 × 7 = 56
But this counts every handshake twice — once from each end.
56 ÷ 2 = 28
A hexagon has 6 vertices. Join every pair of vertices by a straight line.
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Reveal answer
There are C(6,2) = 15 pairs of vertices.
Six of those pairs are sides of the hexagon.
15 − 6 = 9
So there are 9 diagonals.
Look for a simpler object
Sometimes a complicated-looking problem becomes easy if we match each object we want to count with a simpler object.
For example, every subset of 7 objects corresponds to a seven-place yes/no pattern:
in, out, in, in, out, out, in
Each of the 7 positions has 2 choices, so there are 27 subsets.
A grid has 4 vertical lines and 5 horizontal lines.
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Reveal answer
Choose 2 of the 4 vertical lines:
C(4,2) = 6
Choose 2 of the 5 horizontal lines:
C(5,2) = 10
Every pair of vertical choices can be combined with every pair of horizontal choices:
6 × 10 = 60
So the grid contains 60 rectangles.
Your counting toolbox
There is rarely just one way to solve a counting problem. Before calculating, ask what structure the problem has.
- List systematically when the number of possibilities is small.
- Multiply when choices happen in stages.
- Use factorials for arrangements.
- Use combinations when order does not matter.
- Use Pascal’s triangle to see relationships between combination numbers.
- Subtract from the total when the opposite is easier.
- Split into cases when possibilities fall into separate types.
- Use symmetry to avoid repeating equivalent work.
- Count twice when two viewpoints reveal the same collection.
The arithmetic is often the easy part. The real skill is finding a way to organise the possibilities so that every object is counted exactly once — or understanding precisely why you have counted it more than once.
You have reached the end of the Counting & Combinatorics course.
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