Adding two set sizes counts the overlap twice. Subtracting it once fixes that, and the whole counting rule is exactly that correction.
Today's hook, If 30 students study music and 20 study art, you cannot say 50 study one or the other. Anyone doing both has been counted twice, and the counting rule is how you take the extra copy back.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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In a class of 25, 14 play soccer and 16 play netball. Add those two numbers. Why is the answer bigger than the class? What does the excess tell you?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Add, then take the overlap back once
Work through the core explanation before applying it.
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Add, then take the overlap back once
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Adding $n(A)$ and $n(B)$ counts anything in both sets twice, once inside each total. Subtracting $n(A \cap B)$ once removes exactly the extra copy, giving $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$ disjoint sets: the last term is 0
Subtract once, not twice
The overlap is counted twice in the sum, so removing one copy is enough. Subtracting twice leaves the union short.
It rearranges four ways
Any three of the four quantities give the fourth. Most exam questions supply three and ask for the missing one.
Disjoint is the special case
When nothing is shared, $n(A \cap B) = 0$ and the rule collapses to plain addition.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$ for any two finite sets.
The subtraction corrects for elements counted once in each set.
The rule rearranges to find any one of the four quantities from the other three.
For disjoint sets the intersection term is zero.
Understand
Concepts
Why the overlap is counted exactly twice in $n(A) + n(B)$.
Why subtracting it once, rather than twice, is the right correction.
Why the rule reduces to simple addition when the sets are disjoint.
Can do
Skills
Establish the rule from a Venn diagram.
Use it to find any one of the four quantities given the other three.
Apply it to worded problems, including finding an overlap that was not given.
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Key terms
Counting ruleThe formula linking the sizes of two sets, their overlap and their union. Like this: with $n(A) = 30$, $n(B) = 20$ and $n(A \cap B) = 8$, the union has $42$.
Double countingCounting the same element more than once, which is what $n(A) + n(B)$ does to the overlap. Like this: a student doing both music and art is inside both totals.
OverlapAnother word for the intersection, the elements shared by both sets. Like this: the 8 students doing both subjects.
Rearranging the ruleUsing the formula to find whichever quantity is missing. Like this: $n(A \cap B) = n(A) + n(B) - n(A \cup B)$.
At least oneThe wording that signals the union, since it means in $A$, in $B$, or in both. Like this: how many study at least one of the two asks for $n(A \cup B)$.
NeitherThe region outside both circles, found by subtracting the union from the universal set. Like this: $n(\xi) - n(A \cup B)$.
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Establishing the rule
Work through the core explanation before applying it.
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Establishing the rule
core concept
Draw two overlapping circles. Every element is in exactly one of three regions inside them: $A$ only, both, or $B$ only.
Now count $n(A) + n(B)$. The $A$ only region is counted once, in $n(A)$. The $B$ only region is counted once, in $n(B)$. But the overlap is counted twice, once in each.
The union should count every region once. So subtract one copy of the overlap: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$. It is a correction rather than a new fact.
Why exactly once. The overlap appears twice in the sum and should appear once in the union, so exactly one copy is surplus. Subtracting twice would leave the union missing the overlap entirely.
Quick check: $n(A) = 12$, $n(B) = 9$, $n(A \cap B) = 4$. What is $n(A \cup B)$?
Adding $n(A)$ and $n(B)$ counts the overlap twice while the union needs it once. Subtracting $n(A \cap B)$ removes exactly the surplus copy, giving $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Pause, copy the three-region diagram, the argument that the overlap is counted twice, and the rule itself, into your book.
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Rearranging to find what is missing
core concept
We just saw where the rule comes from. That raises a question: what if the question gives you the union and asks for the overlap instead? This card answers it → the rule has four quantities, so any three give the fourth.
The rule links $n(A)$, $n(B)$, $n(A \cap B)$ and $n(A \cup B)$. Given any three, rearrange for the fourth.
To find the overlap: $n(A \cap B) = n(A) + n(B) - n(A \cup B)$. With 14 soccer players, 16 netball players and a class of 25 who all play at least one, the overlap is $14 + 16 - 25 = 5$.
That is the standard exam shape: the overlap is the thing not given, recovered from the excess when the two totals are added.
At least one means the union. Wording such as how many play at least one sport is asking for $n(A \cup B)$, not for $n(A) + n(B)$.
Fill the blank: if $n(A) = 14$, $n(B) = 16$ and $n(A \cup B) = 25$, then $n(A \cap B) = $ .
The rule has four quantities and any three give the fourth. Rearranged for the overlap it reads $n(A \cap B) = n(A) + n(B) - n(A \cup B)$. Wording such as at least one signals the union.
Pause, copy the rearrangement for the overlap, the soccer and netball example, and the note about at least one, into your book.
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Using it with a universal set
core concept
We just saw how to recover any missing quantity. That raises a question: what about the people in neither set? This card answers it → they are everything left over once the union is known.
The union counts everyone in at least one set. Anyone in neither is outside both circles, so $n(\text{neither}) = n(\xi) - n(A \cup B)$.
With 80 people, 45 dog owners, 32 cat owners and 15 owning both: the union is $45 + 32 - 15 = 62$, so those owning neither number $80 - 62 = 18$.
That gives a full accounting: 30 dog only, 15 both, 17 cat only, 18 neither, totalling 80. Checking the four regions against $n(\xi)$ confirms the work.
For disjoint sets the rule simplifies. $n(A \cap B) = 0$, so $n(A \cup B) = n(A) + n(B)$. That is not a different rule; it is the same one with a zero term.
Which is NOT a correct rearrangement of the counting rule?
Those in neither set are $n(\xi) - n(A \cup B)$. Adding the four regions must give $n(\xi)$, which checks the work. For disjoint sets the intersection term is zero and the rule becomes plain addition.
Pause, copy the neither calculation, the full four-region accounting, and the disjoint special case, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · FINDING THE UNION
In a group, 30 study music, 20 study art and 8 study both. How many study at least one of the two?
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$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
At least one means the union.
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$= 30 + 20 - 8$
Substitute the three given values.
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$= 42$ students
The 8 doing both were counted twice, so one copy is removed.
PROBLEM 2 · FINDING THE OVERLAP
In a class of 25, every student plays soccer or netball or both. 14 play soccer and 16 play netball. How many play both?
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Everyone plays at least one, so $n(A \cup B) = 25$
The union is the whole class.
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$n(A \cap B) = n(A) + n(B) - n(A \cup B)$
Rearrange for the overlap.
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$= 14 + 16 - 25 = 5$ students
The excess of the sum over the class size is the overlap.
PROBLEM 3 · INCLUDING NEITHER
Of 80 people, 45 own a dog, 32 own a cat and 15 own both. How many own neither?
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$n(A \cup B) = 45 + 32 - 15 = 62$
Find the union first.
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$n(\text{neither}) = 80 - 62$
Everyone not in the union is outside both circles.
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$= 18$ people. Check: $30 + 15 + 17 + 18 = 80$ ✓
The four regions total the universal set.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
If $n(A) = 10$, $n(B) = 7$ and $n(A \cap B) = 3$, find $n(A \cup B)$.
If $n(A) = 20$, $n(B) = 15$ and $n(A \cup B) = 30$, find $n(A \cap B)$.
Two disjoint sets have 9 and 6 elements. How many are in their union?
With $n(\xi) = 50$ and $n(A \cup B) = 38$, how many are in neither?
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Revisit the soccer and netball class
Run the quick drill and copy the summary into your book.
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Revisit the soccer and netball class
At the start you added 14 and 16 and found the total exceeded the class of 25. Explain what that excess represents, and connect it to the minus sign in the counting rule.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. In a group of 60 people, 38 speak French, 27 speak German and 12 speak both. How many speak at least one of the two languages, and how many speak neither? (3 marks)
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ApplyBand 43 marks
Q2. A club has 42 members. 25 play chess, 23 play bridge and every member plays at least one. Find how many play both. (3 marks)
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UnderstandBand 43 marks
Q3. Explain why the counting rule subtracts the overlap exactly once rather than twice, referring to how it is counted in $n(A) + n(B)$. (3 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $10 + 7 - 3 = 14$. Practice 2: $20 + 15 - 30 = 5$. Practice 3: $9 + 6 = 15$. Practice 4: $50 - 38 = 12$.
Q1 (3 marks): $n(F \cup G) = 38 + 27 - 12$ [1]. $= 53$ speak at least one [1]. Neither $= 60 - 53 = 7$ [1].
Q2 (3 marks): Every member plays at least one, so $n(C \cup B) = 42$ [1]. $n(C \cap B) = 25 + 23 - 42$ [1]. $= 6$ members play both [1].
Q3 (3 marks): An element in both sets appears once inside $n(A)$ and once inside $n(B)$, so the sum counts it twice [1]. The union should count every element exactly once [1]. Removing one copy leaves it counted once; removing two would leave it out of the union altogether [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Counting Rule Rush
earn bronze · silver · gold
Apply the counting rule in all four directions and account for the neither region. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.