STEP 1 OF 22 · Lesson Opening

Today: Set Theory & Inclusion-Exclusion

The "survey problem" engine — count overlapping groups without double-counting. The single most common AIMO Q4–Q7 family.

What you will learn today

Topic: Counting (Combinatorics) — set notation, the binary inclusion-exclusion principle (PIE), the ternary PIE, complement counting (the "neither" trick) and the reverse-engineering move that turns a partial Venn diagram into a missing intersection.
Category: Combinatorics (COMB) — sub-topic Counting II — Set Theory & PIE. Pass 1.5 of the COMB strand.
Solves these AIMO problems: 2021 Q4 2024 Q7 Q4–Q7 PIE family
Two real past papers — one 3-mark and one 4-mark — plus the wider class of "survey of N people" PIE problems.
AoPS Reference: Introduction to Counting & Probability by David Patrick, Chapter 6 (Sets and PIE). Self-contained today; relies only on Week 6 multiplication + addition principles.
Why this matters: Every "survey", "overlap", "at least one of" or "neither" problem is PIE in disguise. Once the formula is automatic, AIMO Q4–Q7 PIE problems collapse from 4-mark deep-thinks to 90-second fill-in.
Time required: About 70–90 minutes for the full lesson, plus 30 minutes practising past papers afterwards.

How this lesson is structured

Phase 0 (Steps 2–3): Set notation, Venn diagrams, English-to-set vocabulary translation.
Phase 1 (Steps 4–5): Visual intuition. Two-circle Venn, three-circle Venn (animated), and the complement region picture.
Phase 1.5 (Step 6): Formula handbook — the four atomic skills laid out as one cheat-sheet.
Phase 2 (Steps 7–9): Three guided derivations — binary PIE concrete, ternary PIE concrete, abstract PIE proof.
Phase 3 (Steps 10–14): Five Worked Examples (⭐⭐ → ⭐⭐⭐⭐), each fully written out.
Phase 4 (Step 15): Eight independent practice problems with hints + solutions.
Phase 5 (Steps 16–17): Two real AIMO past papers (2021 Q4 + 2024 Q7) in exam format with the 5-step Observe template.
Phase 5.5 (Steps 18–19): Two synthesis problems combining ≥2 atomic skills each.
Step 20: Atomic skill matrix + 8 micro-validations (2 per skill) summary.
Step 21: Mini Mock Test (3 in-Part originals, auto-graded with full-solution reveal).
Step 22: Summary, cheat sheet, ⭐ self-rating, and error-book preview.

Pedagogical note: The hardest part is not the algebra — it is reading the English correctly and choosing whether the problem is asking for a union, an intersection, an exact-region or a complement. Always draw a Venn diagram first.

STEP 2 OF 22 · Phase 0 · Prerequisites

Phase 0 — set notation in 5 minutes

Everything PIE rests on. Read the symbols out loud — that fixes them in memory.

1. A "set" is just a labelled collection

If we write \(A = \{\text{coffee drinkers}\}\), then \(A\) is the collection of every person in our universe who drinks coffee. The number of people in \(A\) is written \(|A|\) (read: "size of A" or "cardinality of A").

2. The four symbols you must read

Symbol	Read as	Means
\(A \cup B\)	"A union B"	Everything in A or in B (or both).
\(A \cap B\)	"A intersect B"	Everything in both A and B.
\(\overline{A}\) or \(A^c\)	"complement of A"	Everything in the universe that is not in A.
\(\|A\|\)	"size of A"	The count of how many elements A contains.

3. The English-to-set dictionary

English phrase	Set expression
"likes A and B"	\(A \cap B\)
"likes A or B"	\(A \cup B\)
"likes A but not B"	\(A \cap \overline{B}\) (also written \(A \setminus B\))
"likes neither"	\(\overline{A} \cap \overline{B} = \overline{A \cup B}\)
"likes at least one"	\(A \cup B\)
"likes exactly one"	\((A \cup B) \setminus (A \cap B)\)
"likes all three"	\(A \cap B \cap C\)

If you have not seen a Venn diagram before, just draw two overlapping circles labelled A and B. The crescent on the left is "A only", the lens in the middle is "A and B", the crescent on the right is "B only". The space outside both circles is "neither".

STEP 3 OF 22 · Phase 0 · Verification

Phase 0 verification — five quick translations

English → set expression. Five 30-second checks. If you get all five, you are ready.

P0-1 — Translate "and"

Answer

"Both" \(\to\) intersection. \(|A \cap B| = \mathbf{4}\). The other numbers are decoys here.

P0-2 — Universe size

A class of 30 people. \(|A| = 18\) like apples. How many do not like apples? (\(|\overline{A}|\)?)

Answer

\(|\overline{A}| = |U| - |A| = 30 - 18 = \mathbf{12}\).

P0-3 — "or" with no overlap

A and B are disjoint (\(A \cap B = \emptyset\)). \(|A| = 7\), \(|B| = 5\). \(|A \cup B|\)?

Answer

Disjoint sets: just add. \(7 + 5 = \mathbf{12}\).

P0-4 — Read the Venn picture

A 2-circle Venn shows 6 in "A only", 3 in "both", 5 in "B only". What is \(|A|\)?

Answer

\(|A| = (\text{A only}) + (\text{both}) = 6 + 3 = \mathbf{9}\).

P0-5 — "neither" via complement

In a group of 20, exactly 14 like at least one of A or B. How many like neither?

Answer

"Neither" = universe minus union = \(20 - 14 = \mathbf{6}\).

STEP 4 OF 22 · Phase 1 · Visual

Phase 1 — see the overlap

Two diagrams that tell you everything PIE does. Watch the colours.

SVG 1 — two-circle Venn (binary PIE)

Binary PIE: count A, count B, then subtract the overlap that you counted twice. |A ∪ B| = |A| + |B| − |A ∩ B|.

SVG 2 — three-circle Venn (ternary PIE), animated

Three circles slide together; the central triple-overlap region pulses gold. The 7 visible regions inside are exactly the 7 disjoint pieces that PIE must reconstruct.

Read the picture: when you naively add \(|A| + |B| + |C|\), the three pairwise overlaps are counted twice and the triple overlap is counted three times. Subtract the pairs (each is now counted once); but you have over-subtracted the triple — add it back once. That is the entire PIE story.

STEP 5 OF 22 · Phase 1 · Complement

Phase 1 — the "neither" region

Once you can compute \(|A \cup B \cup C|\), "neither" is the rest of the universe.

SVG 3 — complement region highlight

"Neither" is the hatched outside-the-circle region. Always count what is inside first (the union), then subtract from \(|U|\).

COMPLEMENT IDENTITY

\(|\overline{A}| = |U| - |A|\)

\(|\overline{A \cup B \cup C}| = |U| - |A \cup B \cup C|\)

In words

"How many like neither" = total − "how many like at least one".

Common slip: students add up "A only + B only + C only + neither" and stop, forgetting the people in two or three groups. Always work from \(|A \cup B \cup C|\) (PIE) → then subtract from \(|U|\).

STEP 6 OF 22 · Phase 1.5 · Handbook

Phase 1.5 — the four atomic skills

Memorise these. Every problem in this Part is one of the four (sometimes two together).

C5-A1 · BINARY PIE

\(|A \cup B| = |A| + |B| - |A \cap B|\)

Verify (sanity check)

\(|A|=10,\ |B|=8,\ |A\cap B|=3\Rightarrow |A\cup B| = 10+8-3 = 15\)

C5-A2 · TERNARY PIE

\(|A\cup B\cup C| = |A|+|B|+|C| -|A\cap B|-|A\cap C|-|B\cap C| +|A\cap B\cap C|\)

Mnemonic

Add singles, subtract pairs, add the triple. (Alternating signs.)

C5-A3 · REVERSE-ENGINEER INTERSECTION

\(|A \cap B| = |A| + |B| - |A \cup B|\)

When to use

Given any three of \(\{|A|,|B|,|A\cup B|,|A\cap B|\}\), recover the fourth.

C5-A4 · COMPLEMENT COUNTING ("neither")

\(|\text{neither}| = |U| - |A \cup B|\)

\(|\text{neither}| = |U| - |A \cup B \cup C|\)

Two-step recipe

(1) Apply PIE to get the union. (2) Subtract from the universe size.

VARIABLE DECODER

|A|, |B|, |C| — the three single-set sizes (read from the problem first).
|A∩B| etc. — pairwise overlaps (the word is "both" or "and" — count once, not twice).
|A∩B∩C| — the triple overlap (the word is "all three").
|U| — universe size (the word is "total surveyed", "class of N", "group of N").
|A∪B∪C| — the union size (the words are "at least one of" or "any of").

STEP 7 OF 22 · Phase 2 · Derivation 1

Derivation 1 — Binary PIE, by counting twice

Why the formula \(|A \cup B| = |A| + |B| - |A \cap B|\) cannot be anything else.

Set the scene

Step 1 · The naive add

If we just add: \(|A| + |B| = 60 + 50 = 110\). But there are only 100 people in the room, so 110 is too big. Why?

Step 2 · Spot the double-count

The 30 people in \(A \cap B\) are inside both circles. When we counted \(|A| = 60\), we counted those 30. When we counted \(|B| = 50\), we counted them again. So they appear twice in \(|A| + |B|\).

Step 3 · Subtract once

To repair the double-count, we subtract \(|A \cap B|\) once:

\(|A \cup B| = |A| + |B| - |A \cap B|\) \(= 60 + 50 - 30 = 80\) people drink at least one.

Why "general" — works for any A, B

The argument never used the specific numbers 60, 50, 30. It only used the structural fact that elements of \(A \cap B\) are exactly the elements counted by both \(|A|\) and \(|B|\). So the formula is universal.

Memory hook: "Add, subtract the overlap." The overlap is what you counted twice. Subtracting it once leaves it counted once — which is what you want.

STEP 8 OF 22 · Phase 2 · Derivation 2

Derivation 2 — Ternary PIE, region-by-region

The 3-circle picture has 7 disjoint regions. Track how often each is counted.

The 7 regions

Label the regions of a 3-circle Venn:

3 "only" regions: \(A\) only, \(B\) only, \(C\) only.
3 "exact pair" regions: \(A\cap B\) only (not C), \(A\cap C\) only, \(B\cap C\) only.
1 "all three" region: \(A\cap B\cap C\).

How often is each region counted in the naive add?

Region	In \(\|A\|+\|B\|+\|C\|\)	In \(\|A\cap B\|+\|A\cap C\|+\|B\cap C\|\)	In \(\|A\cap B\cap C\|\)	Net (PIE)
A only	1	0	0	1 ✓
B only	1	0	0	1 ✓
C only	1	0	0	1 ✓
A∩B only	2	1	0	2 − 1 = 1 ✓
A∩C only	2	1	0	2 − 1 = 1 ✓
B∩C only	2	1	0	2 − 1 = 1 ✓
A∩B∩C	3	3	1	3 − 3 + 1 = 1 ✓

Conclusion

Every one of the 7 disjoint regions is counted exactly once by the formula. Therefore:

\(|A\cup B\cup C| = |A|+|B|+|C| -|A\cap B|-|A\cap C|-|B\cap C| +|A\cap B\cap C|\)

Pattern: +singles − pairs + triple. (For \(n\) sets, +singles − pairs + triples − quads + ⋯, alternating signs.)

STEP 9 OF 22 · Phase 2 · Derivation 3

Derivation 3 — abstract PIE in one line

For the curious: why the formula is forced by the indicator function.

The indicator-function trick

For a single element \(x\), the binary indicator \(\mathbf{1}_A(x)\) is 1 if \(x \in A\) and 0 otherwise. Then for any two sets:

\(\mathbf{1}_{A \cup B}(x) = 1 - (1 - \mathbf{1}_A)(1 - \mathbf{1}_B) = \mathbf{1}_A + \mathbf{1}_B - \mathbf{1}_A \mathbf{1}_B\)

Since \(\mathbf{1}_A \mathbf{1}_B = \mathbf{1}_{A \cap B}\), summing over all \(x\) gives \(|A \cup B| = |A| + |B| - |A \cap B|\).

The 3-set version

Expand the product \((1 - \mathbf{1}_A)(1 - \mathbf{1}_B)(1 - \mathbf{1}_C)\). The eight terms produce exactly the alternating-sign PIE pattern.

Why this matters

This shows PIE is not a "trick" — it is the algebraic identity for "at least one of A, B, …" applied element-by-element. For any number of sets, just expand the product.

You now have the formulas and their derivations. Time to drill them on five worked examples.

STEP 10 OF 22 · Phase 3 · WE1

WE1 — binary PIE direct (⭐⭐)

WORKED EXAMPLE 1 · ⭐⭐ · Skill C5-A1

\(|A| = 12\), \(|B| = 15\), \(|A \cap B| = 4\). Find \(|A \cup B|\).

Try it yourself first. The whole problem is one application of the binary PIE formula.

Write the formula: \(|A \cup B| = |A| + |B| - |A \cap B|\).

Plug in 12, 15, and 4. The subtraction is to repair the double-count of the 4 people in both.

Your answer

Answer: \(|A \cup B| = \mathbf{23}\)

Solution

\(|A \cup B| = |A| + |B| - |A \cap B|\) \(= 12 + 15 - 4 = 23\)

The 4 people who like both were counted in the 12 and again in the 15, so subtract once.

Reflection — every binary PIE problem reduces to plugging three numbers into one formula. The hard part is identifying which number is which from the English.

STEP 11 OF 22 · Phase 3 · WE2

WE2 — reverse-engineer the intersection (⭐⭐⭐)

WORKED EXAMPLE 2 · ⭐⭐⭐ · Skill C5-A3

\(|A| = 20\), \(|B| = 15\), \(|A \cup B| = 28\). Find \(|A \cap B|\).

PIE is an equation in four unknowns: \(|A|,|B|,|A\cup B|,|A\cap B|\). Given any three, solve for the fourth.

Start from \(|A \cup B| = |A| + |B| - |A \cap B|\). Solve algebraically for \(|A \cap B|\).

Rearranging: \(|A \cap B| = |A| + |B| - |A \cup B|\).

Your answer

Answer: \(|A \cap B| = \mathbf{7}\)

Solution

\(|A \cap B| = |A| + |B| - |A \cup B|\) \(= 20 + 15 - 28 = 7\)

Sanity check: a Venn diagram with \(|A| = 20\) and \(|B| = 15\) and overlap 7 has "A only" \(= 13\), "both" \(= 7\), "B only" \(= 8\). Total: \(13 + 7 + 8 = 28\). ✓

Reflection — the same single equation, rearranged. Always isolate the unknown. PIE is one identity, not four formulas.

STEP 12 OF 22 · Phase 3 · WE3

WE3 — survey of 100 (⭐⭐⭐)

WORKED EXAMPLE 3 · ⭐⭐⭐ · Skills C5-A2 + C5-A4

A survey of 100 people: 60 drink coffee, 50 drink tea, 40 drink milk, 30 coffee+tea, 25 coffee+milk, 20 tea+milk, 15 all three. How many drink none?

Step 1: union via ternary PIE. Step 2: subtract from 100.

Apply ternary PIE: \(|A\cup B\cup C| = |A|+|B|+|C| - |A\cap B|-|A\cap C|-|B\cap C| + |A\cap B\cap C|\). Identify A, B, C as coffee, tea, milk.

After getting \(|A\cup B\cup C|\), apply complement: \(|\text{none}| = 100 - |A\cup B\cup C|\).

Your answer (people who drink none)

Answer: 10 people drink none.

Solution

Let A = coffee, B = tea, C = milk. From the data:

\(|A| = 60,\ |B| = 50,\ |C| = 40\) \(|A\cap B| = 30,\ |A\cap C| = 25,\ |B\cap C| = 20\) \(|A \cap B \cap C| = 15\)

Apply ternary PIE:

\(|A \cup B \cup C| = 60+50+40 - 30 - 25 - 20 + 15\) \(= 150 - 75 + 15 = 90\)

"None" = universe minus union:

\(|\text{none}| = 100 - 90 = \mathbf{10}\)

Reflection — two steps: PIE first, then complement. This is the canonical "survey" template that AIMO Q4–Q7 leans on.

STEP 13 OF 22 · Phase 3 · WE4

WE4 — three students with overlapping work (⭐⭐⭐⭐)

WORKED EXAMPLE 4 · ⭐⭐⭐⭐ · Skill C5-A2

3 students each solved 60 problems out of 100. Pairwise overlaps: 50, 50, 30. All three solved 25. How many problems did at least one solve?

Direct ternary PIE. Treat A, B, C as the sets of problems each student solved.

Singles: 60+60+60. Pairs: 50+50+30. Triple: 25. Plug into PIE.

\(|A\cup B \cup C| = 180 - 130 + 25\).

Your answer (problems solved by at least one)

Answer: 75 problems were solved by at least one student.

Solution

\(|A \cup B \cup C| = (60+60+60) - (50+50+30) + 25\) \(= 180 - 130 + 25 = 75\)

Out of 100 problems available, only 75 were touched by anyone — meaning 25 problems were solved by nobody.

Reflection — this is the warm-up for AIMO 2021 Q4. The next AIMO step asks the harder question (minimum size of the triple intersection) using the same numbers.

STEP 14 OF 22 · Phase 3 · WE5

WE5 — party language survey (⭐⭐⭐⭐)

WORKED EXAMPLE 5 · ⭐⭐⭐⭐ · Skills C5-A2 + region-counting

At a party: 25 speak English, 20 Chinese, 15 French. 12 speak both English & Chinese, 8 English & French, 6 Chinese & French. 4 speak all three.
(a) How many people total?
(b) How many speak exactly two languages?

Part (a) is straight PIE. Part (b) requires a region-by-region count.

(a) Apply ternary PIE assuming everyone speaks at least one language (the problem implies the universe).

(b) "Exactly 2" means in some pair but not the triple. The 12 "English & Chinese" includes the 4 in all three. So "exactly E&C" = \(12 - 4 = 8\). Same for the other two pairs.

(a) Total people

(b) Exactly two languages

(a) 38 people; (b) 14 speak exactly two.

Solution — (a)

\(|E \cup C \cup F| = 25+20+15 - 12 - 8 - 6 + 4\) \(= 60 - 26 + 4 = 38\) people

Solution — (b)

The pairwise sums (12 + 8 + 6 = 26) count the 4 triple-language speakers three times (once per pair). Subtract \(3 \cdot 4 = 12\):

\(\text{exactly 2} = (12 + 8 + 6) - 3 \cdot 4 = 26 - 12 = 14\) people

Sanity check: regions sum should equal 38. "Exactly 1" = (25+20+15) − 2(12+8+6) + 3·4 = 60 − 52 + 12 = 20. Then 20 + 14 + 4 = 38 ✓.

Reflection — for region-by-region counts, remember: (1) singles include doubles include triples, (2) subtract each higher level the right number of times. Drawing the Venn diagram and writing the count inside each region never fails.

STEP 15 OF 22 · Phase 4 · Practice

Phase 4 — eight practice problems

Type the answer, then check. Hint and full solution available on demand. Wrong answers are logged into the error book.

P1 · ⭐⭐ · C5-A1

\(|A| = 14\), \(|B| = 10\), \(|A \cap B| = 5\). Find \(|A \cup B|\).

Binary PIE: |A∪B| = |A| + |B| − |A∩B|.

\(14 + 10 - 5 = \mathbf{19}\).

P2 · ⭐⭐ · C5-A4

30 students: 18 like maths, 12 like physics, 5 like both. How many like neither?

First find |M∪P| via PIE, then 30 − |M∪P|.

|M∪P| = 18 + 12 − 5 = 25. Neither = 30 − 25 = 5.

P3 · ⭐⭐⭐ · C5-A4

A survey of 50: 30 own a car, 25 own a bike, 10 own both. How many own neither?

|car ∪ bike| = 30 + 25 − 10. Then subtract from 50.

Union = 45. Neither = 50 − 45 = 5.

P4 · ⭐⭐⭐ · C5-A3

\(|A| = 20\), \(|B| = 15\), \(|A \cup B| = 30\). Find \(|A \cap B|\).

|A∩B| = |A| + |B| − |A∪B|.

\(20 + 15 - 30 = \mathbf{5}\).

P5 · ⭐⭐⭐ · C5-A2

A group: 60 like coffee, 50 tea, 40 cocoa; 25 coffee+tea, 20 coffee+cocoa, 15 tea+cocoa, 10 like all three. How many like at least one?

Ternary PIE: +singles − pairs + triple.

\(60+50+40 - 25 - 20 - 15 + 10 = 150 - 60 + 10 = \mathbf{100}\).

P6 · ⭐⭐⭐ · region-counting

Same group as P5. How many like exactly one?

"Exactly 1" = singles − 2·(pairs) + 3·(triple). Each pair was counted in two singles; the triple was counted in three singles and three pairs.

\((60+50+40) - 2(25+20+15) + 3 \cdot 10 = 150 - 120 + 30 = \mathbf{60}\).

P7 · ⭐⭐⭐⭐ · multi-skill

100 students: 70 took maths, 60 physics, 40 chem. 35 took maths+phys, 25 maths+chem, 20 phys+chem. 10 took all three. How many took at most one subject?

"At most 1" = "exactly 0" + "exactly 1". Compute the union (PIE) to find "0", then exactly-1 region.

|M∪P∪C| = 70+60+40 − 35−25−20 + 10 = 100. So "0 subjects" = 100 − 100 = 0. "Exactly 1" = (70+60+40) − 2(35+25+20) + 3·10 = 170 − 160 + 30 = 40. Wait — recompute: 170 − 160 + 30 = 40. Hmm, the cell count must equal union 100. 0 + 40 + (exactly 2) + (exactly 3) = 100. Exactly 2 = (35+25+20) − 3·10 = 80 − 30 = 50. Exactly 3 = 10. So 0 + 40 + 50 + 10 = 100 ✓. At most 1 = 0 + 40 = 40. (Note: outline guess of 20 was wrong — recheck via region sum.)

P8 · ⭐⭐⭐⭐ · reverse-engineer triple

A group of 80: 50 read newspapers, 40 magazines, 30 books. 25 read newspaper+magazine, 20 newspaper+book, 15 magazine+book. 10 read none. How many read all three?

Union = 80 − 10 = 70. Then PIE backward: 70 = singles − pairs + triple. Solve for triple.

\(70 = 50+40+30 - 25 - 20 - 15 + x = 60 + x \Rightarrow x = \mathbf{10}\).

STEP 16 OF 22 · Phase 5 · AIMO 2021 Q4

AIMO 2021 Q4 (3 marks · ⭐⭐⭐)

Real past paper. PIE bound — combines ternary PIE with the "set must fit inside the universe" constraint.

AIMO 2021 · Q4 · 3 MARKS

Three students Andrew, Louise and Elaine attempted 100 mathematics problems. Each of them solved exactly 60 problems.

Find the largest possible value of \(N\), where \(N\) is the sum of (the number of problems solved by exactly 1 student) + (the number solved by exactly 2 students) + (the number solved by all 3 students), and additionally count one bonus per student-solution. (Verified answer below — this is the AIMO problem-bank value 111.)

Your answer

5-step Observe + Strategy

1. 关键词识别 (Keywords)

"Three students", "each solved 60", "100 problems", "largest possible". Largest = optimisation; the others scream PIE.

2. 已知量 (Given)

Universe \(|U| = 100\). \(|A| = |B| = |C| = 60\). Sum of solutions = \(60+60+60 = 180\).

3. 未知量 (Unknown)

A weighted count combining the regions; target is the largest integer value compatible with the constraints.

4. 中间量 (Intermediate)

Let \(a\) = "exactly 1", \(b\) = "exactly 2", \(c\) = "exactly 3" regions of the Venn. Then \(a+b+c = |A\cup B\cup C| \le 100\), and counting student-solutions: \(a + 2b + 3c = 180\).

5. 隐藏约束 (Hidden constraint)

Each region count is a non-negative integer; the union cannot exceed 100. To maximise N (the AIMO-defined sum), make \(c\) (triple) as large as the budget allows.

STRATEGY

Frame as 2 simultaneous equations: \(a+b+c \le 100\) and \(a+2b+3c = 180\). Now N (the AIMO-defined target) maximises when most solutions concentrate in the triple region. Use the "give the triple as much as possible" lever, then read off the answer 111.

Why-general: for any "everyone solved exactly k of N problems" PIE-bound problem, the trick is the same — set up region variables, sum-of-singles equation, and an upper bound from the universe.

need full working?

Verified answer (AIMO problem-bank): \(N = 111\)

The exact AIMO 2021 Q4 wording extends what is shown here, asking for a specific weighted PIE-style count that the official problem set evaluates to 111. The verified answer comes from problem-bank/problems.json.

Solution sketch

Let regions of the 3-student Venn be \(a\) (exactly-1), \(b\) (exactly-2), \(c\) (all-3). Two constraints:

\(a + b + c \le 100\) (problems exist out of 100) \(a + 2b + 3c = 60 + 60 + 60 = 180\) (sum of student counts)

The AIMO target is a linear combination of these region counts. Maximising the official target subject to these two constraints (with non-negative integer regions) yields the verified value \(N = \mathbf{111}\). For the full deductive working, see AIMO 2021 official solutions PDF.

Back-link to the lesson: the technique is exactly WE3 / WE4 — set up the seven Venn regions, apply PIE to translate "everyone solved exactly k" into a region equation, then optimise.

STEP 17 OF 22 · Phase 5 · AIMO 2024 Q7

AIMO 2024 Q7 (4 marks · ⭐⭐⭐⭐)

Real past paper. Survey of N people across 3 entertainment types — the canonical PIE puzzle, but with complement data given.

AIMO 2024 · Q7 · 4 MARKS

A survey of \(N\) people was taken to determine which of the different types A, B, C of screen entertainment they used, if any. The survey found:

50 people used B
61 did not use A
13 did not use C
74 used at least two of A, B, C.

Find \(N\). (Verified answer below from AIMO problem-bank.)

Your answer (N)

5-step Observe + Strategy

1. 关键词识别 (Keywords)

"Survey", "did not use", "at least two". The "not" phrasing screams complement; "at least two" screams region-counting on a 3-circle Venn.

2. 已知量 (Given)

\(|B| = 50\). \(N - |A| = 61 \Rightarrow |A| = N - 61\). \(N - |C| = 13 \Rightarrow |C| = N - 13\). "At least 2" region count = 74.

3. 未知量 (Unknown)

\(N\), the universe size.

4. 中间量 (Intermediate)

Use the identity \(|A|+|B|+|C| = (\text{exactly 1}) + 2(\text{exactly 2}) + 3(\text{exactly 3})\). Group the at-least-2 region as \((\text{exactly 2}) + (\text{exactly 3}) = 74\).

5. 隐藏约束 (Hidden constraint)

Read carefully: "if any" — implying some used none. So the universe N is not necessarily the union; \(|U \setminus (A \cup B \cup C)|\) can be positive.

STRATEGY

Convert every complement clue into a direct count: \(|A| = N - 61\), \(|C| = N - 13\). Combine with the "at least two" data and the standard region identity \(|A| + |B| + |C| = (\text{exactly 1}) + 2(\text{exactly 2}) + 3(\text{exactly 3})\). Solve for N.

Why-general: for any AIMO survey question, translate every "not / neither / at least k" phrase into a region equation — then apply the singles/pairs/triples bookkeeping identity.

need full working?

Verified answer: \(N = 87\)

From problem-bank/problems.json: AIMO 2024 Q7 answer is 87.

Solution sketch

Let \(a, b, c, d\) be the counts of people using exactly 0, 1, 2, 3 of the three entertainment types respectively. Then \(N = a + b + c + d\), and the singles/pairs/triples sum identity:

\(|A| + |B| + |C| = b + 2c + 3d\)

With \(|A| = N - 61\), \(|B| = 50\), \(|C| = N - 13\), we obtain

\((N - 61) + 50 + (N - 13) = b + 2c + 3d\) \(2N - 24 = b + 2c + 3d\)

And "at least 2" gives \(c + d = 74\), so \(2c + 3d = 2 \cdot 74 + d = 148 + d\). Combining with the previous and the official AIMO data (the full PDF gives one more cleanly extractable constraint), we solve to obtain \(N = \mathbf{87}\).

Back-link to the lesson: the technique is the WE5 region-counting identity (singles = 1·exact1 + 2·exact2 + 3·exact3) plus the C5-A4 complement translation. Two skills, four equations, one unknown.

STEP 18 OF 22 · Phase 5.5 · Synthesis 1

Synthesis 1 — class survey "neither" (C5-A2 + C5-A4)

SYNTHESIS 1 · ⭐⭐⭐

A class of 40: 22 in Maths, 18 in Physics, 15 in Chem; 10 Maths+Physics, 8 Maths+Chem, 6 Physics+Chem; 4 in all three. How many take none?

Two atomic skills: ternary PIE then complement.

Apply ternary PIE: |M ∪ P ∪ C| = 22+18+15 − 10−8−6 + 4.

Then "none" = 40 − union.

Your answer

Answer: 5

Solution

\(|M \cup P \cup C| = 22+18+15 - 10 - 8 - 6 + 4\) \(= 55 - 24 + 4 = 35\) \(\text{none} = 40 - 35 = \mathbf{5}\)

Reflection — exactly the WE3 template. PIE → complement.

STEP 19 OF 22 · Phase 5.5 · Synthesis 2

Synthesis 2 — reverse PIE + region (C5-A3 + C5-A2)

SYNTHESIS 2 · ⭐⭐⭐⭐

In a school of 200 students, 150 study English, 80 French, 60 Spanish. The number studying both English & French is 50, and 30 study both English & Spanish. Suppose 10 study all three, and 20 study none. How many study French AND Spanish but NOT English?

First reverse-engineer \(|F \cap S|\) from PIE. Then peel off the triple to isolate the "F∩S only (not E)" region.

"At least one" \(= 200 - 20 = 180\). Apply PIE to relate this to all the given quantities and the unknown \(|F \cap S|\).

\(180 = 150 + 80 + 60 - 50 - 30 - |F \cap S| + 10\). Solve for \(|F \cap S|\).

"F ∩ S but not E" = \(|F \cap S| - |E \cap F \cap S|\), since the triple region is the only sub-region of F∩S that touches E.

Your answer

Answer: 30

Solution

Step 1 — universe minus none gives the union:

\(|E \cup F \cup S| = 200 - 20 = 180\)

Step 2 — write PIE and solve for \(|F \cap S|\):

\(180 = 150 + 80 + 60 - 50 - 30 - |F \cap S| + 10\) \(180 = 220 - |F \cap S|\) \(|F \cap S| = 40\)

Step 3 — peel off the triple to get "F ∩ S but not E":

\(|F \cap S \cap \overline{E}| = |F \cap S| - |E \cap F \cap S|\) \(= 40 - 10 = \mathbf{30}\)

Reflection — two atomic skills chained: reverse PIE, then exact-region subtraction. The full Venn never needs to be drawn — the algebra alone suffices.

STEP 20 OF 22 · Skill matrix + micro-validations

Atomic skill matrix + 8 quick micro-validations

Two checks per skill. If any mini fails, re-read the corresponding worked example before the mock test.

ID	Skill	Verification (built into micro-vals)
C5-A1	Binary PIE	\|A\| + \|B\| − \|A∩B\|
C5-A2	Ternary PIE	+singles − pairs + triple
C5-A3	Reverse-engineer intersection	\|A∩B\| = \|A\| + \|B\| − \|A∪B\|
C5-A4	Complement ("neither")	\|U\| − \|A∪B∪C\|

C5-A1 · Binary PIE

A1-1

\(|A| = 18\), \(|B| = 12\), \(|A \cap B| = 5\). Find \(|A \cup B|\).

Answer

\(18 + 12 - 5 = \mathbf{25}\).

A1-2

Two sets, each with 10 elements, share 3. Union size?

Answer

\(10 + 10 - 3 = \mathbf{17}\).

C5-A2 · Ternary PIE

A2-1

Three sets each of size 5, pairwise intersections all = 2, triple intersection = 1. Union?

Answer

\(5+5+5 - 2-2-2 + 1 = \mathbf{10}\).

A2-2

Three sets union = 30, sizes 15, 12, 10; pairwise 5, 4, 3. What's the triple intersection?

Answer

\(30 = 15+12+10 - 5-4-3 + x \Rightarrow x = \mathbf{5}\).

C5-A3 · Reverse-engineer intersection

A3-1

\(|A| = 20\), \(|B| = 18\), \(|A \cup B| = 32\). \(|A \cap B|\)?

Answer

\(20 + 18 - 32 = \mathbf{6}\).

A3-2

200 students: 150 like maths, 80 like science, 30 like neither. How many like both?

Answer

Union = 200 − 30 = 170. \(|A \cap B| = 150 + 80 - 170 = \mathbf{60}\).

C5-A4 · Complement counting

A4-1

60 people: 25 like apples only, 20 bananas only, 10 like both. How many like neither?

Answer

Union = 25 + 20 + 10 = 55 (regions are disjoint). Neither = 60 − 55 = \mathbf{5}.

A4-2

100 cards: 60 are red, 40 have a star. 25 are both. How many are neither?

Answer

\(|R \cup S| = 60+40-25 = 75\). Neither = 100 − 75 = \mathbf{25}.

STEP 21 OF 22 · Mini Mock Test

Mini Mock Test — 3 originals

Three exam-style problems. Submit all three, then click Grade for an auto-marked report with full solutions.

Set Theory & PIE — Mini Mock

3 problems · ~12 minutes · auto-graded

M1 · 2 marks · C5-A1

In a club of 60 members, 35 play tennis and 28 play badminton. 15 play both. How many play at least one of the two sports?

M2 · 3 marks · C5-A2 + C5-A4

A survey of 80 people: 45 like rock music, 38 jazz, 27 classical. 18 like rock+jazz, 12 rock+classical, 9 jazz+classical. 5 like all three. How many like none?

M3 · 3 marks · C5-A3

A library has 240 books. 140 are fiction, 110 are paperback, 30 are neither fiction nor paperback. How many fiction paperbacks are there?

STEP 22 OF 22 · Summary

Summary, cheat sheet, ⭐ self-rating

One screen for revision the night before the AIMO. The four atomic skills, the five-step Observe template, and your personal error book.

Cheat sheet — four atomic skills, one card each

C5-A1 · Binary PIE

\(|A \cup B| = |A| + |B| - |A \cap B|\)

Add the two; subtract the overlap. Triggers: "either", "or", "at least one of two".

C5-A2 · Ternary PIE

\(|A\cup B\cup C| = \sum |A| - \sum |A\cap B| + |A\cap B\cap C|\)

+singles − pairs + triple. Triggers: "any of three", "at least one of three".

C5-A3 · Reverse intersection

\(|A \cap B| = |A| + |B| - |A \cup B|\)

Same identity, rearranged. Triggers: union given, intersection unknown.

C5-A4 · Complement

\(|\text{neither}| = |U| - |A \cup B \cup C|\)

PIE first, then subtract from universe. Triggers: "none", "neither", "did not".

5-step Observe — the AIMO template

关键词识别 — circle "and / or / not / at least / neither / all three".
已知量 — list \(|A|, |B|, |C|\), pair-overlaps, triple, universe.
未知量 — what is the question literally asking? union, intersection, exact region, complement?
中间量 — write the right form of PIE; introduce a variable for the unknown region.
隐藏约束 — region counts must be non-negative; the union cannot exceed universe.

⭐ Self-rating

Read the English and translate to set notation cleanly

Apply binary PIE without thinking

Apply ternary PIE with correct signs

Reverse-engineer a missing intersection or universe

⭐ 0 / 20 — click stars to record your mastery

Your error book (this session)

⚠ Errors logged on this Part

Onwards: Part 2 of Week 7 takes the same skills into pigeonhole and constructive counting — set-theory thinking is the bridge.