Principle of Inclusion–Exclusion
When you count how many people study Maths or Physics, you can't just add — those who study both get counted twice. The Principle of Inclusion–Exclusion is the systematic fix: add individual sets, subtract pairwise overlaps, add back triple overlaps, and so on. It's the engine behind Venn diagrams and a surprisingly powerful combinatorial tool.
In a class of 30 students, 18 study Maths and 15 study Physics. Without using any formula — do you think it's possible that all 15 Physics students also study Maths? Is it possible that none of them do? If you're told exactly 8 study both, how many study at least one subject? Write your reasoning.
Every inclusion–exclusion problem asks you to do one of two things: find the union given the individual sets and their overlaps, or find an intersection or missing piece given the union and other information.
The entire lesson turns on one idea: when you add two sets, you double-count their overlap. So you subtract it once. For three sets, you over-subtract the triple overlap, so you add it back. The formula alternates: add singles, subtract pairs, add triples, subtract quadruples, and so on.
Key facts
- $|A \cup B| = |A| + |B| - |A \cap B|$
- $|A \cup B \cup C| = |A|+|B|+|C| - |A\cap B| - |B\cap C| - |A\cap C| + |A\cap B\cap C|$
- Union = OR; intersection = AND; neither = outside all sets
Concepts
- Why adding two sets double-counts the intersection
- Why the triple intersection is subtracted once too many in the three-set formula and must be added back
- How Venn diagrams make inclusion–exclusion calculations visual and error-free
Skills
- Apply the two-set formula to solve word problems
- Apply the three-set formula with all pairwise and triple intersections
- Draw and label Venn diagrams to verify results
Suppose you want to count how many elements are in $A$ or $B$ (at least one of the two sets). If you simply add $|A| + |B|$, every element in both sets gets counted twice. So you subtract the overlap once:
Rearranging, you can also find the intersection if you know the union: $|A \cap B| = |A| + |B| - |A \cup B|$. And the number in neither set is $|U| - |A \cup B|$.
The overlap region $A \cap B$ is subtracted once so it is counted exactly once in the union.
|A B| = |A| + |B| - |A B|. Always subtract the intersection.; Rearrangement: |A B| = |A| + |B| - |A B| (find overlap if you know the rest)
Pause — copy the two-set inclusion-exclusion formula into your book: $|A \cup B| = |A| + |B| - |A \cap B|$; rearrangement: $|A \cap B| = |A| + |B| - |A \cup B|$.
Quick check: In a class of 30 students, 18 study Maths, 15 study Physics, and 8 study both. How many study at least one of these subjects?
We just saw that $|A \cup B| = |A| + |B| - |A \cap B|$ corrects for the double-counting of elements in both sets. That raises a question: when three sets overlap, elements in all three are counted in three different pairwise intersections — how does inclusion-exclusion extend to remove all excess counting? This card answers it → $|A \cup B \cup C| = |A|+|B|+|C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|$.
For three overlapping sets, the double-counting problem is more complex. Each element in exactly two sets gets counted twice (once for each set it belongs to); each element in all three sets gets counted three times. Applying inclusion–exclusion carefully:
Why add back $|A \cap B \cap C|$? When you subtract the three pairwise intersections, you subtract the triple intersection three times — once in each pairwise term. But it should only be subtracted twice (since it was added three times). Net: subtract $3 - 3 + 1 = 1$ time total. The triple intersection ends up counted exactly once. ✓
Memory aid: singles (add), pairs (subtract), triple (add). Alternating signs, starting positive.
|A B C| = |A|+|B|+|C| - |A B| - |B C| - |A C| + |A B C|; Each element in exactly one set is counted 1-0+0=1 ✓
Pause — copy the three-set inclusion-exclusion formula into your book: $|A \cup B \cup C| = |A|+|B|+|C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|$.
Did you get this? True or false: in the three-set inclusion–exclusion formula, $|A \cap B \cap C|$ is added (not subtracted).
Worked examples · 3 in a row, reveal as you go
In a class of 30 students, 18 study Maths ($M$), 15 study Physics ($P$), and 8 study both. How many study at least one of these subjects? How many study neither?
In a survey, 40 people like tea, 35 like coffee, and 55 like at least one of the two. How many like both? How many like neither if the survey had 60 people?
Find $|A \cup B \cup C|$ if $|A|=20$, $|B|=25$, $|C|=18$, $|A \cap B|=8$, $|B \cap C|=6$, $|A \cap C|=5$, $|A \cap B \cap C|=2$.
Fill the gap: If $|A|=30$, $|B|=40$, and $|A\cap B|=12$, then $|A\cup B| = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for three sets, the number of elements in at least one set equals $|A|+|B|+|C|-|A\cap B|-|B\cap C|-|A\cap C|$ (with no other correction needed).
Activities · practice with the ideas
In a survey, 40 people like tea, 35 like coffee, and 20 like both. How many like at least one? How many like neither if 60 were surveyed?
In a group of 50 people, 30 speak French and 25 speak Spanish. If 45 speak at least one language, how many speak both?
Find $|A \cup B \cup C|$ if $|A|=20$, $|B|=25$, $|C|=18$, $|A \cap B|=8$, $|B \cap C|=6$, $|A \cap C|=5$, $|A \cap B \cap C|=2$.
How many integers from 1 to 100 are divisible by 2 or 3? (Hint: $|A| = \lfloor 100/2 \rfloor$, $|B| = \lfloor 100/3 \rfloor$, $|A\cap B| = \lfloor 100/6 \rfloor$.)
Explain why the formula for $|A \cup B \cup C|$ has a positive $|A \cap B \cap C|$ term. What would go wrong if you omitted it?
Odd one out: Which statement about the three-set inclusion–exclusion formula is NOT correct?
Earlier you were asked: if 18 study Maths, 15 study Physics, and 8 study both — how many study at least one?
$|M \cup P| = 18 + 15 - 8 = 25$ students study at least one subject. This is why you can't just add 18 and 15: the 8 students who study both would be counted twice, giving the wrong answer of 33.
Both students in the hook were partially right: at least $18+15-30=3$ study both (if all 30 study at least one) and at most $\min(18,15)=15$ study both. Since exactly 8 study both, the union is exactly 25.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. In a survey, 40 people like tea, 35 like coffee, and 20 like both. How many like at least one? (2 marks)
Q2. Find $|A \cup B \cup C|$ if $|A|=20$, $|B|=25$, $|C|=18$, $|A \cap B|=8$, $|B \cap C|=6$, $|A \cap C|=5$, $|A \cap B \cap C|=2$. (3 marks)
Q3. How many integers from 1 to 120 are divisible by 2 or 3 or 5? Use inclusion–exclusion. Let $A$, $B$, $C$ be the sets of multiples of 2, 3, 5 respectively. (3 marks)
Comprehensive answers (click to reveal)
Activities: 1. $40+35-20=55$; neither $=60-55=5$ · 2. $|F\cap S|=30+25-45=10$ · 3. $20+25+18-8-6-5+2=46$ · 4. $50+33-16=67$ · 5. Without $+|A\cap B\cap C|$, the triple overlap is subtracted three times instead of two, so elements in all three sets are counted zero times instead of one — the union is under-counted by $|A\cap B\cap C|$.
Q1 (2 marks): $|T\cup C|=40+35-20=55$ [2].
Q2 (3 marks): Singles $=63$; pairs $=8+6+5=19$; triple $=2$. $|A\cup B\cup C|=63-19+2=46$ [3].
Q3 (3 marks): $|A|=60$, $|B|=40$, $|C|=24$ [0.5]. $|A\cap B|=20$, $|A\cap C|=12$, $|B\cap C|=8$ [1]. $|A\cap B\cap C|=4$ [0.5]. $|A\cup B\cup C|=60+40+24-20-12-8+4=88$ [1].
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