Problems involving sets
Master problems involving sets for Year 11 Specialist Mathematics in Queensland (QCAA). A Venn diagram pictures how two or three sets overlap, so you can organise and count a survey at a glance.
You will learn the four set operations — union, intersection, complement and difference — read set-builder notation, and pull the "only", "exactly" and "neither" numbers from a diagram. It is the set-reasoning foundation for counting and probability.
Theory
Problems involving sets use Venn diagrams and set notation to organise a collection of objects in Year 11 Specialist Mathematics (QCAA, Queensland). This page covers the four set operations — union, intersection, complement and difference — set-builder notation, and how to read the only, exactly and neither regions of a diagram.
A set is a collection of distinct objects called its elements. Write \(x\in A\) for "\(x\) is an element of \(A\)", and \(|A|\) for the number of elements in \(A\) (its cardinality). The universal set \(\xi\) contains every object under discussion; a subset \(A\subseteq\xi\) has all its elements inside \(\xi\).
The four operations combine sets. The union \(A\cup B\) collects every element in \(A\) or \(B\) (each once). The intersection \(A\cap B\) keeps only elements in both. The complement \(A'\) holds the elements of \(\xi\) that are not in \(A\). The difference \(A\setminus B\) keeps the elements of \(A\) that are not in \(B\).
Set-builder notation describes a set by a rule instead of a list: \(A=\{x : x \text{ is a factor of } 12\}\) means "all \(x\) such that \(x\) is a factor of \(12\)". You still list and count the elements the rule produces.
On a Venn diagram the wording matters. Only \(A\) means the part of \(A\) outside every other circle; exactly one adds the separate only-regions; and neither is everything outside all circles, \(|\xi|-|A\cup B|\).
The complement takes every element of the universal set that is not in \(A\):
The difference removes the shared part from \(A\):
For two sets, exactly one adds the two only-regions:
How to read a set problem
- Name the sets and the universal set \(\xi\); note each given cardinality (\(|A|\), \(|A\cap B|\), and so on).
- Fill the overlap first. Place the innermost count (\(A\cap B\), or the all-three region) before anything else, then subtract to find each only-region.
- Match the wording to a region: "only" is one lobe, "exactly one/two" sums the matching regions, and "neither" is outside every circle.
- Count or combine the required regions, checking that all regions add back to \(|\xi|\).
Union — every element in \(A\) or \(B\), each listed once:
| \(A\cup B\) | \(=\) | \(\{1,2,3,4,5,6,8,10\}\) |
| \(|A\cup B|\) | \(=\) | \(8\) |
Intersection — the elements in both sets:
| \(A\cap B\) | \(=\) | \(\{2,4\}\) |
| \(|A\cap B|\) | \(=\) | \(2\) |
Complement — the elements of \(\xi\) not in \(A\):
| \(A'\) | \(=\) | \(\{1,3,5,7,9\}\) |
| \(|A'|\) | \(=\) | \(5\) |
Difference — the elements of \(A\) with those in \(B\) removed:
| \(A\setminus B\) | \(=\) | \(\{6,8,10\}\) |
| \(|A\setminus B|\) | \(=\) | \(3\) |
\(A\cup B=\{1,2,3,4,5,6,8,10\}\), \(A\cap B=\{2,4\}\), \(A'=\{1,3,5,7,9\}\), \(A\setminus B=\{6,8,10\}\).
Apply the rule — list the multiples of \(3\) up to \(12\):
| \(A\) | \(=\) | \(\{3,6,9,12\}\) |
| \(|A|\) | \(=\) | \(4\) |
Complement — the elements of \(\xi\) that are not multiples of \(3\):
| \(A'\) | \(=\) | \(\{1,2,4,5,7,8,10,11\}\) |
| \(|A'|\) | \(=\) | \(|\xi| - |A|\) |
| \(=\) | \(12 - 4\) | |
| \(=\) | \(8\) |
\(A=\{3,6,9,12\}\), so \(A'=\{1,2,4,5,7,8,10,11\}\) and \(|A'|=8\).
Subtract the overlap of \(7\) to get each only-region:
| \(\text{yoga only}\) | \(=\) | \(24 - 7\) |
| \(=\) | \(17\) | |
| \(\text{Pilates only}\) | \(=\) | \(18 - 7\) |
| \(=\) | \(11\) |
Exactly one adds the two only-regions:
| \(\text{exactly one}\) | \(=\) | \(17 + 11\) |
| \(=\) | \(28\) |
Neither is the total minus the union of the three inside regions:
| \(|Y\cup P|\) | \(=\) | \(17 + 7 + 11\) |
| \(=\) | \(35\) | |
| \(\text{neither}\) | \(=\) | \(40 - 35\) |
| \(=\) | \(5\) |
\(28\) members do exactly one activity, and \(5\) do neither.
Each pair overlap minus the all-three region gives its exactly-two part:
| \(A\cap B \text{ only}\) | \(=\) | \(10 - 3\) |
| \(=\) | \(7\) | |
| \(A\cap C \text{ only}\) | \(=\) | \(8 - 3\) |
| \(=\) | \(5\) | |
| \(B\cap C \text{ only}\) | \(=\) | \(6 - 3\) |
| \(=\) | \(3\) |
Add the three exactly-two regions:
| \(\text{exactly two}\) | \(=\) | \(7 + 5 + 3\) |
| \(=\) | \(15\) |
\(15\) elements lie in exactly two of the sets.
Common pitfalls
Frequently asked questions
What is the difference between union and intersection?
The union \(A\cup B\) collects every element in \(A\) or \(B\) (each once); the intersection \(A\cap B\) keeps only the elements that are in both sets.
What does the complement \(A'\) mean?
It is every element of the universal set \(\xi\) that is not in \(A\). Its size is \(|A'|=|\xi|-|A|\), so you must know \(\xi\) first.
What does set-builder notation \(\{x : \dots\}\) mean?
It describes a set by a rule rather than a list. \(\{x : x \text{ is a factor of } 12\}\) reads "all \(x\) such that \(x\) is a factor of \(12\)", giving \(\{1,2,3,4,6,12\}\).
How do you find 'exactly one' on a Venn diagram?
Subtract the overlap from each set to get the two only-regions, then add them: \((|A|-|A\cap B|)+(|B|-|A\cap B|)\).
What does 'neither' mean on a Venn diagram?
It is everything outside all the circles — the complement of the union, counted as \(|\xi|-|A\cup B|\).
Is \(A\setminus B\) the same as \(B\setminus A\)?
No. \(A\setminus B\) keeps the elements of \(A\) not in \(B\); \(B\setminus A\) keeps the elements of \(B\) not in \(A\). They are usually different sets.