Union — A ∪ B ∪ C
Everything in A, B, or C — all regions shaded
Intersection — A ∩ B ∩ C
All 7 regions of a 3-set Venn diagram labelled
A only — A − (B ∪ C)
Elements in A but in neither B nor C
Complement — Ā
Everything in 𝕌 outside A (B, C may exist)
#
Cardinality of Sets
📌
Definition
Cardinality of set A, written n(A) or |A|, is the number of distinct elements in A.
Example: A = {a,b,c,d} → n(A) = 4
Example: A = {a,b,c,d} → n(A) = 4
Key Rules — Three Sets
1
Count each element only once. {1,1,2,3} has cardinality 3.
2
2-set union: n(A∪B) = n(A) + n(B) − n(A∩B)
3
3-set inclusion-exclusion:
n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)
n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)
4
Complement: n(Ā) = n(𝕌) − n(A)
5
Disjoint sets (no overlap): n(A∩B) = 0, so n(A∪B) = n(A)+n(B)
6
A-only region: n(A only) = n(A) − n(A∩B) − n(A∩C) + n(A∩B∩C)
7
Elements in none: n(𝕌) − n(A∪B∪C)
Work through each example. Click Next Example when ready.
Example 1 of 3Basic Cardinality of a Single Set
A = {3, 7, 11, 15, 19}
Find n(A).
👀Elements: 3, 7, 11, 15, 19
🔢Count: 1,2,3,4,5 — all distinct
Answer
n(A) = 5
Example 2 of 3n(A∪B) using 2-set formula
A={1,2,3,4,5}B={4,5,6,7}
📊n(A)=5, n(B)=4
🔍A∩B = {4,5} → n(A∩B)=2
🧮5+4−2 = 7
Answer
n(A∪B) = 7
Example 3 of 33-Set Inclusion-Exclusion
n(A)=20n(B)=18n(C)=15
n(A∩B)=5n(B∩C)=4n(A∩C)=6n(A∩B∩C)=2
➕Sum: 20+18+15 = 53
➖Subtract pairwise: 53−(5+4+6) = 38
➕Add triple: 38+2 = 40
Answer
n(A∪B∪C) = 40
All questions involve three sets A, B and C. Answer each step to unlock the next — the Venn diagram shows what you're solving!
1
Question 1 of 8
A={1,2,3,4}, B={3,4,5,6}, C={4,6,7,8}. Find n(A∪B∪C).
💡Hint: n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)
Step 1: Count n(A), n(B), n(C) from the sets and add them = ?
✓ Final Answer
n(A∪B∪C) = 8
All 7 regions displayed: A-only {1,2}, B-only {5}, C-only {7,8}, intersections {3}, {6}, ∅, triple {4}
2
Question 2 of 8
A={1,2,3,4}, B={3,4,5,6}, C={4,6,7,8}. Find n(A∩B∩C).
💡Hint: Find elements common to all three sets.
A∩B = elements in both A and B?
✓ Final Answer
n(A∩B∩C) = 1 — only {4}
Same intersection layout as Set Operations, now filled with the actual values from A, B and C.
3
Question 3 of 8
n(A)=10, n(B)=8, n(C)=6, n(A∩B)=3, n(B∩C)=2, n(A∩C)=2, n(A∩B∩C)=1. Find n(A∪B∪C).
💡Hint: Apply inclusion-exclusion: sum − pairwise + triple.
Step 1: n(A)+n(B)+n(C) = ?
✓ Final Answer
n(A∪B∪C) = 18
All regions by cardinality: 6+4+3 = 13 only; 2+1+1 = 4 pairs; 1 triple = 18 total
4
Question 4 of 8
A={1,2,3,4}, B={3,4,5,6}, C={4,6,7,8}. Find n(A only) — elements exclusively in A, not in B or C.
💡Hint: A only = A − (B∪C). Remove anything A shares with B or C.
Step 1: Count B ∪ C from the sets or Venn = ?
✓ Final Answer
n(A only) = 2 — elements {1, 2}
A only (A − (B∪C)) = {1, 2} shown in left pink region
5
Question 5 of 8
𝕌={1..15}, A=multiples of 3, B=multiples of 5, C=multiples of 2, all up to 15. Find n(Ā) relative to 𝕌.
💡Hint: n(Ā) = n(𝕌) − n(A). List multiples of 3 up to 15 first.
A = {3,6,9,12,15}. What is n(A)?
✓ Final Answer
n(Ā) = 10
n(Ā) = 15 − 5 = 10 elements outside A
🔥 Challenge Questions
6
🔥 Challenge · Q6/8
In a class of 50: n(A)=20, n(B)=25, n(C)=18, n(A∩B)=8, n(B∩C)=6, n(A∩C)=5, n(A∩B∩C)=3. Find n(A∪B∪C).
💡Hint: Apply 3-set inclusion-exclusion step by step.
Step 1: n(A)+n(B)+n(C) = ?
✓ Final Answer
n(A∪B∪C) = 47 · Neither = 3
Challenge: 10+14+10 = 34 only; 5+3+2 = 10 pairs; 3 triple = 47 total
7
🔥 Challenge · Q7/8
n(A∪B∪C)=45, n(A)=22, n(B)=20, n(C)=18, n(A∩B)=7, n(B∩C)=5, n(A∩C)=6. Find n(A∩B∩C).
💡Hint: Rearrange: n(A∩B∩C) = n(A∪B∪C) − n(A) − n(B) − n(C) + n(A∩B) + n(B∩C) + n(A∩C)
Step 1: n(A)+n(B)+n(C) = ?
✓ Final Answer
n(A∩B∩C) = 3
Same intersection layout as Set Operations, filled with the derived region values for this question.
8
🔥 Challenge · Q8/8
n(𝕌)=60, n(A)=25, n(B)=22, n(C)=18, n(A∩B)=9, n(B∩C)=7, n(A∩C)=6, n(A∩B∩C)=4. How many are in NONE of A, B, C?
💡Hint: Find n(A∪B∪C) first, then None = n(𝕌) − n(A∪B∪C).
Step 1: n(A)+n(B)+n(C) = ?
✓ Final Answer
n(A∪B∪C)=47 · None = 13
Challenge: 12+10+9 = 31 only; 5+3+2 = 10 pairs; 4 triple = 45 union → 60−45 = 15 none
📝 Written Solution
Solution builds here
Select correct options to build the written solution automatically.