At a glance. One lesson of about 60 minutes. It follows the set laws and assumes the two-set counting rule. Students apply the three-set rule, justify it by counting how many times each region is counted, solve for a missing total, find the number in exactly two sets, and predict the pattern for four sets.
Calculator: no calculator. No calculator is needed. The sums are small, and the point of the lesson is the structure of the rule, not the arithmetic.
What students will learn
Know
- n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C).
- Each element is counted exactly once: 1, then 2 − 1, then 3 − 3 + 1.
- Elements in exactly two sets = sum of pairs − 3 n(A ∩ B ∩ C).
Understand
- A rule is justified by showing that every case is counted once, not by showing that it works on one example.
- The terms alternate in sign because each step corrects the over-count of the step before.
- The same structure extends to four sets: 15 terms with alternating signs.
Do
- Use the rule to find a union or a number in none of the sets.
- Use the rule backwards to find an unknown intersection total.
- Count elements in exactly one or two sets from totals.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: applying the three-set rule, forwards and backwards (questions 2 to 6; question 1 checks the four-set multiplicity).
- Criterion B. Investigating patterns: the figure and the what-if. Students find how many times each region is counted at each step, and predict the pattern for four sets.
- Criterion C. Communicating: justifying the rule in a short, ordered argument (worked example 3) with correct set notation.
- Criterion D. Applying mathematics in real-life contexts: the clubs hook, where adding three reported totals gives a wrong announcement.
The one thing to do with the figure
Ask for a prediction at every step. With survey 1 on step 1, ask: “Which regions are counted more than once?” Then press step 2 and ask which region now has a count of 0 and why.
Then choose survey 4 and ask why the last step changes nothing. Students should say the centre region holds 0 elements.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the head teacher's total. Collect predictions. |
| 12 min | The figure with predictions; the key idea; settle the hook. |
| 10 min | Warning card and worked examples; pairs finish the faded example. |
| 5 min | Question 1, the check. If many choose A, list the six pairs on the board and count again. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 8. |
The answers
| Warm-up | A′ ∩ B′; 24; 3. |
|---|---|
| Hook | Too high: elements in two or three clubs are counted more than once. Taking off the pairs makes it too low, because the centre is taken off as often as it was added. |
| Worked examples | 77; 5; each element is counted once (1, 2 − 1, 3 − 3 + 1). |
| Faded example | 40. |
| 1. Multiplicity with four sets | C. 4 − 6 + 4. A copies the three-set centre; B is the count after the last step; D stops after the totals. |
| 2. The three-set rule | 55. 74 − 22 + 3. |
| 3. Elements in none of three sets | 22. 120 − 98. |
| 4. Elements in exactly two sets | 25. 37 − 3 × 4. |
| 5. Using the rule backwards | 20. 57 + 3 − 40. |
| 6. Exactly two from total memberships | 35. 120 − 85. |
| 7. Terms in the four-set rule | 15. 4 + 6 + 4 + 1. |
| 8. At least two of three sets | B. 24 − 2 × 3. A is exactly two only; C takes the centre off once; D is the pair totals. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A, B or D in question 1 | Copied the three-set answer for the centre, counted past the last step, or stopped after the totals. | List the six pairs AB, AC, AD, BC, BD, CD and tick each one the element is in. |
| 52, 49 or 74 in question 2 | Stopped after the pairs, subtracted the triple, or gave the totals only. | Say the three steps aloud: total, minus pairs, plus triple. |
| 98, 30 or 38 in question 3 | Gave the union, left out the triple, or subtracted it. | Find the union first, then ask what the question wants. |
| 37, 33 or 29 in question 4 | Gave the pair totals, or removed the centre once or twice. | Draw the three pair regions and shade the centre, which sits in all three. |
| 17 or 14 in question 5 | Left out the triple or subtracted it. | Write the rule with the numbers in, then solve. |
| 50, 85 or 65 in question 6 | Gave exactly one, at most two, or the extra memberships. | Use two equations: people, and memberships. |
| 10, 14 or 16 in question 7 | Missed the triples or the quadruple, or counted the empty choice. | List the groups: 4, 6, 4, 1. |
| A, C or D in question 8 | Gave exactly two, took the centre off once, or added the pair totals. | Draw the three pair regions and the centre, and ask how many times the centre is inside the 24. |
Questions worth asking
- Why must the triple be added back? The centre is added three times and taken off three times, so it is counted 0 times, and needs one more count.
- What happens to the rule if the three sets are pairwise disjoint? All pair totals and the triple are 0, so the union is the sum of the totals.
- How would you check the rule on a real survey? Count the regions directly and compare, as the figure does.
- What would the four-set rule look like? Totals, minus pairs, plus triples, minus the quadruple: signs alternate and there are 15 terms.
Differentiation
A way in for anyone stuck
- Give a three-set Venn template and ask students to write the multiplicity of each region after each step.
- Provide the three steps as a numbered frame: total, minus pairs, plus triple.
- For the exactly-two count, draw the three pair regions and shade the centre in a different colour.
Stretch
- Find the number of elements in exactly one set, in terms of the totals, the pair totals and the triple.
- Prove that the four-set rule counts an element in all four sets once, by evaluating 4 − 6 + 4 − 1.
Students learning in English as an additional language
“At least one” means one or more, so it is the union. “Exactly two” means in two sets and not in the third. “None” means in no set, so it is the complement of the union. Say “n of A” for n(A).
| Word | Plain meaning | Example |
|---|---|---|
| union | all elements in at least one of the sets | club members in robotics or choir or both |
| intersection | elements in all the sets named | members of both robotics and choir |
| at least one | one or more of the sets | a student in one, two or three clubs |
| exactly two | in two of the sets and not in the third | in robotics and choir only |
| multiplicity | how many times an element has been counted so far | an element counted twice so far has multiplicity 2 |
| pairwise disjoint | no two of the sets share an element | three clubs with no student in two of them |
Sentence starters for explaining aloud or in writing:
- An element in exactly two sets is counted … times after step 1, and … times after step 2.
- The triple must be added because …
- The rule counts every element once because …
Two things not to say
“Add the three totals and take off the overlaps.” Say which overlaps and what happens to the centre.
“Check the rule on one survey, so it is proved.” One example works for one survey; the argument covers every survey.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- Applyn(X) = 20, n(Y) = 16, n(Z) = 12, n(X ∩ Y) = 6, n(X ∩ Z) = 5, n(Y ∩ Z) = 4 and n(X ∩ Y ∩ Z) = 2. Find the union.
48 − 15 + 2 = 35. - Earlier work (Standard Unit 4)n(P) = 28, n(Q) = 19 and n(P ∩ Q) = 11. Find n(P ∪ Q).
36.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextIn a class of 40, 18 study French, 14 Spanish and 10 German; 6 take French and Spanish, 4 French and German, 3 Spanish and German, and 1 takes all three. How many study none?
Union = 42 − 13 + 1 = 30, so 10 study none. - Missing valuen(A ∪ B ∪ C) = 50, n(A) = 25, n(B) = 22, n(C) = 18, the pair totals sum to 20. Find n(A ∩ B ∩ C).
50 = 65 − 20 + t, so t = 5 (20 ≥ 3 × 5, and pair totals 8, 6 and 6 give regions 16, 13, 11, 3, 1, 1 and 5, which add to 50).
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningShow that the number of elements in exactly one set is n(A) + n(B) + n(C) − 2(sum of the pair totals) + 3 n(A ∩ B ∩ C).
Count how many times each kind of element is counted. Exactly one set: 1. Exactly two sets: 2 − 2 × 1 = 0. All three: 3 − 2 × 3 + 3 = 0. So only the elements in exactly one set remain. - Four setsAn element is in exactly three of four sets. How many times does the four-set rule count it?
3 − 3 + 1 = 1: it is in 3 totals, 3 pairs and 1 triple, and in no quadruple.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
The figure works with the keyboard (tab to a slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.
No calculator is needed. A printed three-set template helps with the multiplicities.