Chapter 3
Exercise 3.2 Solutions
Step-by-step solved set operations, De Morgan's Law verifications, Venn diagrams, and Inclusion-Exclusion word problems.
Question 1
Tabular & Venn
Consider the universal set U = {x: x is a multiple of 2 and 0 < x ≤ 30}.
Let A = {x: x is a multiple of 6} and B = {x: x is a multiple of 8}.
(i) List all elements of sets A and B in tabular form.
(ii) Find A ∩ B.
(iii) Draw a Venn diagram.
Let A = {x: x is a multiple of 6} and B = {x: x is a multiple of 8}.
(i) List all elements of sets A and B in tabular form.
(ii) Find A ∩ B.
(iii) Draw a Venn diagram.
Solution
Write out the universal set:
U = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30}.
(i) Tabular forms:
Set A consists of multiples of 6 in U: A = {6, 12, 18, 24, 30}.
Set B consists of multiples of 8 in U: B = {8, 16, 24}.
Set A consists of multiples of 6 in U: A = {6, 12, 18, 24, 30}.
Set B consists of multiples of 8 in U: B = {8, 16, 24}.
(ii) Intersection A ∩ B:
Find common elements between A and B:
A ∩ B = {6, 12, 18, 24, 30} ∩ {8, 16, 24} = {24}.
Find common elements between A and B:
A ∩ B = {6, 12, 18, 24, 30} ∩ {8, 16, 24} = {24}.
(iii) Venn Diagram Layout:
- Exclusive to A: {6, 12, 18, 30}
- Exclusive to B: {8, 16}
- Intersection A ∩ B: {24}
- Universal elements outside A or B: {2, 4, 10, 14, 20, 22, 26, 28}.
- Exclusive to A: {6, 12, 18, 30}
- Exclusive to B: {8, 16}
- Intersection A ∩ B: {24}
- Universal elements outside A or B: {2, 4, 10, 14, 20, 22, 26, 28}.
Question 2
Set Operations
Let U = {x: x is an integer and 0 < x ≤ 150}.
Let G = {x: x = 2m for integer m, where elements belong to U} and H = {x: x is a perfect square}.
(i) List all elements of sets G and H in tabular form.
(ii) Find G ∪ H.
(iii) Find G ∩ H.
Let G = {x: x = 2m for integer m, where elements belong to U} and H = {x: x is a perfect square}.
(i) List all elements of sets G and H in tabular form.
(ii) Find G ∪ H.
(iii) Find G ∩ H.
Solution
Universal set: U = {1, 2, 3, ..., 150}.
(i) Tabular forms:
Powers of 2 in U (for integers m ≥ 0):
G = {1, 2, 4, 8, 16, 32, 64, 128}.
Perfect squares in U (from 12 to 122):
H = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}.
Powers of 2 in U (for integers m ≥ 0):
G = {1, 2, 4, 8, 16, 32, 64, 128}.
Perfect squares in U (from 12 to 122):
H = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}.
(ii) Union G ∪ H:
Combine all elements (without duplicates):
G ∪ H = {1, 2, 4, 8, 9, 16, 25, 32, 36, 49, 64, 81, 100, 121, 128, 144}.
Combine all elements (without duplicates):
G ∪ H = {1, 2, 4, 8, 9, 16, 25, 32, 36, 49, 64, 81, 100, 121, 128, 144}.
(iii) Intersection G ∩ H:
Find elements common to both sets:
G ∩ H = {1, 4, 16, 64} (Note: these are powers of 2 with even exponents, i.e., perfect squares).
Find elements common to both sets:
G ∩ H = {1, 4, 16, 64} (Note: these are powers of 2 with even exponents, i.e., perfect squares).
Question 3
Primes & Divisors
Let P = {x: x is a prime number and 0 < x < 20} and Q = {x: x is a divisor of 210 and 0 < x < 20}.
Find: (i) P ∩ Q and (ii) P ∪ Q.
Find: (i) P ∩ Q and (ii) P ∪ Q.
Solution
List elements of P in tabular form (primes less than 20):
P = {2, 3, 5, 7, 11, 13, 17, 19}.
P = {2, 3, 5, 7, 11, 13, 17, 19}.
List elements of Q in tabular form (divisors of 210 less than 20):
The divisors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, ...
Q = {1, 2, 3, 5, 6, 7, 10, 14, 15}.
The divisors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, ...
Q = {1, 2, 3, 5, 6, 7, 10, 14, 15}.
(i) Intersection P ∩ Q:
P ∩ Q = {2, 3, 5, 7} (Prime factors of 210).
P ∩ Q = {2, 3, 5, 7} (Prime factors of 210).
(ii) Union P ∪ Q:
Combine elements from both sets:
P ∪ Q = {1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19}.
Combine elements from both sets:
P ∪ Q = {1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19}.
Question 4
Commutative Properties
Verify the commutative properties of union and intersection for the following pairs of sets:
(i) A = {1, 2, 3, 4, 5}, B = {4, 6, 8, 10}
(ii) Natural Numbers (N) and Integers (Z)
(iii) A = {x ∈ R | x2 > 0}, B = R
(i) A = {1, 2, 3, 4, 5}, B = {4, 6, 8, 10}
(ii) Natural Numbers (N) and Integers (Z)
(iii) A = {x ∈ R | x2 > 0}, B = R
(i) A = {1, 2, 3, 4, 5}, B = {4, 6, 8, 10}
Solution
Commutative Property of Union (A ∪ B = B ∪ A):
LHS = A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10}.
RHS = B ∪ A = {1, 2, 3, 4, 5, 6, 8, 10}.
Since LHS = RHS, Union is commutative.
LHS = A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10}.
RHS = B ∪ A = {1, 2, 3, 4, 5, 6, 8, 10}.
Since LHS = RHS, Union is commutative.
Commutative Property of Intersection (A ∩ B = B ∩ A):
LHS = A ∩ B = {4}.
RHS = B ∩ A = {4}.
Since LHS = RHS, Intersection is commutative.
LHS = A ∩ B = {4}.
RHS = B ∩ A = {4}.
Since LHS = RHS, Intersection is commutative.
(ii) N (Natural Numbers) and Z (Integers)
Solution
Since every natural number is an integer, N ⊂ Z.
Commutative Property of Union (N ∪ Z = Z ∪ N):
LHS = N ∪ Z = Z.
RHS = Z ∪ N = Z.
LHS = RHS = Z.
LHS = N ∪ Z = Z.
RHS = Z ∪ N = Z.
LHS = RHS = Z.
Commutative Property of Intersection (N ∩ Z = Z ∩ N):
LHS = N ∩ Z = N.
RHS = Z ∩ N = N.
LHS = RHS = N.
LHS = N ∩ Z = N.
RHS = Z ∩ N = N.
LHS = RHS = N.
(iii) A = {x ∈ R | x2 > 0}, B = R
Solution
Observe that x2 > 0 is true for all real numbers except 0. Thus, A = R - {0}.
Clearly, A ⊂ B.
Clearly, A ⊂ B.
Commutative Property of Union:
LHS = A ∪ B = B = R.
RHS = B ∪ A = B = R.
LHS = RHS = R.
LHS = A ∪ B = B = R.
RHS = B ∪ A = B = R.
LHS = RHS = R.
Commutative Property of Intersection:
LHS = A ∩ B = A = R - {0}.
RHS = B ∩ A = A = R - {0}.
LHS = RHS = A.
LHS = A ∩ B = A = R - {0}.
RHS = B ∩ A = A = R - {0}.
LHS = RHS = A.
Question 5
De Morgan's Laws
Let U = {a, b, c, d, e, f, g, h, i, j}.
Let A = {a, b, c, d, g, h} and B = {c, d, e, f, j}.
Verify De Morgan's Laws for these sets.
Let A = {a, b, c, d, g, h} and B = {c, d, e, f, j}.
Verify De Morgan's Laws for these sets.
Solution
De Morgan's Laws state:
Law 1: (A ∪ B)' = A' ∩ B'
Law 2: (A ∩ B)' = A' ∪ B'
Law 1: (A ∪ B)' = A' ∩ B'
Law 2: (A ∩ B)' = A' ∪ B'
Compute complements first:
A' = U - A = {e, f, i, j}
B' = U - B = {a, b, g, h, i}
A' = U - A = {e, f, i, j}
B' = U - B = {a, b, g, h, i}
Verification of Law 1:
- LHS: Find A ∪ B first:
A ∪ B = {a, b, c, d, e, f, g, h, j}.
Then (A ∪ B)' = U - (A ∪ B) = {i}.
- RHS: Find A' ∩ B':
A' ∩ B' = {e, f, i, j} ∩ {a, b, g, h, i} = {i}.
Since LHS = RHS = {i}, Law 1 is verified.
- LHS: Find A ∪ B first:
A ∪ B = {a, b, c, d, e, f, g, h, j}.
Then (A ∪ B)' = U - (A ∪ B) = {i}.
- RHS: Find A' ∩ B':
A' ∩ B' = {e, f, i, j} ∩ {a, b, g, h, i} = {i}.
Since LHS = RHS = {i}, Law 1 is verified.
Verification of Law 2:
- LHS: Find A ∩ B first:
A ∩ B = {c, d}.
Then (A ∩ B)' = U - (A ∩ B) = {a, b, e, f, g, h, i, j}.
- RHS: Find A' ∪ B':
A' ∪ B' = {e, f, i, j} ∪ {a, b, g, h, i} = {a, b, e, f, g, h, i, j}.
Since LHS = RHS, Law 2 is verified.
- LHS: Find A ∩ B first:
A ∩ B = {c, d}.
Then (A ∩ B)' = U - (A ∩ B) = {a, b, e, f, g, h, i, j}.
- RHS: Find A' ∪ B':
A' ∪ B' = {e, f, i, j} ∪ {a, b, g, h, i} = {a, b, e, f, g, h, i, j}.
Since LHS = RHS, Law 2 is verified.
Question 6
Set Identities
If U = {1, 2, 3, ..., 20} and A = {1, 3, 5, ..., 19}, verify:
(i) A ∪ A' = U
(ii) A ∩ U = A
(iii) A ∩ A' = ∅
(i) A ∪ A' = U
(ii) A ∩ U = A
(iii) A ∩ A' = ∅
Solution
Determine set A' (even integers from 1 to 20):
A' = U - A = {2, 4, 6, ..., 20}.
A' = U - A = {2, 4, 6, ..., 20}.
(i) Verify A ∪ A' = U:
LHS = {1, 3, 5, ..., 19} ∪ {2, 4, 6, ..., 20} = {1, 2, 3, ..., 20} = U. (Verified)
LHS = {1, 3, 5, ..., 19} ∪ {2, 4, 6, ..., 20} = {1, 2, 3, ..., 20} = U. (Verified)
(ii) Verify A ∩ U = A:
LHS = {1, 3, 5, ..., 19} ∩ {1, 2, 3, ..., 20} = {1, 3, 5, ..., 19} = A. (Verified)
LHS = {1, 3, 5, ..., 19} ∩ {1, 2, 3, ..., 20} = {1, 3, 5, ..., 19} = A. (Verified)
(iii) Verify A ∩ A' = ∅:
LHS = {1, 3, 5, ..., 19} ∩ {2, 4, 6, ..., 20} = { } = ∅. (Verified)
LHS = {1, 3, 5, ..., 19} ∩ {2, 4, 6, ..., 20} = { } = ∅. (Verified)
Question 7
2-Set Word Problem
In a class of 55 students, 34 like to play cricket and 30 like to play hockey. Also, each student likes to play at least one of the two games. How many students like to play both games?
Solution
Let C represent cricket players and H represent hockey players.
Given quantities:
Total students n(C ∪ H) = 55 (since everyone plays at least one game).
Cricket players n(C) = 34.
Hockey players n(H) = 30.
Total students n(C ∪ H) = 55 (since everyone plays at least one game).
Cricket players n(C) = 34.
Hockey players n(H) = 30.
Apply the Principle of Inclusion-Exclusion for two sets:
n(C ∪ H) = n(C) + n(H) - n(C ∩ H)
55 = 34 + 30 - n(C ∩ H)
55 = 64 - n(C ∩ H)
n(C ∩ H) = 64 - 55 = 9
Conclusion: 9 students like to play both games.
Question 8
3-Set Word Problem
In a group of 500 employees, 250 can speak Urdu, 150 can speak English, 50 can speak Punjabi, 40 can speak both Urdu and English, 30 can speak both English and Punjabi, and 10 can speak both Urdu and Punjabi. How many can speak all three languages?
Solution
Let U, E, and P represent Urdu, English, and Punjabi speakers respectively.
Given data:
Total employees n(U ∪ E ∪ P) = 500.
n(U) = 250, n(E) = 150, n(P) = 50.
Overlaps of two languages:
n(U ∩ E) = 40, n(E ∩ P) = 30, n(U ∩ P) = 10.
Total employees n(U ∪ E ∪ P) = 500.
n(U) = 250, n(E) = 150, n(P) = 50.
Overlaps of two languages:
n(U ∩ E) = 40, n(E ∩ P) = 30, n(U ∩ P) = 10.
Apply the Principle of Inclusion-Exclusion for three sets:
n(U ∪ E ∪ P) = n(U) + n(E) + n(P) - n(U ∩ E) - n(E ∩ P) - n(U ∩ P) + n(U ∩ E ∩ P)
n(U ∪ E ∪ P) = n(U) + n(E) + n(P) - n(U ∩ E) - n(E ∩ P) - n(U ∩ P) + n(U ∩ E ∩ P)
Substitute values into the equation:
500 = 250 + 150 + 50 - 40 - 30 - 10 + n(U ∩ E ∩ P)
500 = 450 - 80 + n(U ∩ E ∩ P)
500 = 370 + n(U ∩ E ∩ P)
n(U ∩ E ∩ P) = 500 - 370 = 130
Conclusion: 130 employees speak all three languages. (Note: This is the textbook solution; logically, the intersection of three sets cannot exceed any of the double overlaps, representing a standard printing error in the curriculum problem.)
Question 9
Shirt & Cap Overlaps
In a sports event, 19 people wear blue shirts, 15 wear green shirts, 3 wear blue and green shirts, 4 wear a cap and blue shirts, and 2 wear a cap and green shirts. The total number of people wearing either a blue or green shirt or a cap is 34. How many people are wearing caps?
Solution
Let B represent blue shirts, G represent green shirts, and C represent caps.
Given values:
n(B) = 19, n(G) = 15, n(C) = x (what we want to find).
Double overlaps: n(B ∩ G) = 3, n(B ∩ C) = 4, n(G ∩ C) = 2.
Triple overlap: n(B ∩ G ∩ C) = 0 (no one wears both green and blue shirts simultaneously).
Total union: n(B ∪ G ∪ C) = 34.
n(B) = 19, n(G) = 15, n(C) = x (what we want to find).
Double overlaps: n(B ∩ G) = 3, n(B ∩ C) = 4, n(G ∩ C) = 2.
Triple overlap: n(B ∩ G ∩ C) = 0 (no one wears both green and blue shirts simultaneously).
Total union: n(B ∪ G ∪ C) = 34.
Apply the Principle of Inclusion-Exclusion:
n(B ∪ G ∪ C) = n(B) + n(G) + n(C) - n(B ∩ G) - n(B ∩ C) - n(G ∩ C) + n(B ∩ G ∩ C)
n(B ∪ G ∪ C) = n(B) + n(G) + n(C) - n(B ∩ G) - n(B ∩ C) - n(G ∩ C) + n(B ∩ G ∩ C)
Substitute values:
34 = 19 + 15 + x - 3 - 4 - 2 + 0
34 = 34 + x - 9
34 = 25 + x
x = 34 - 25 = 9
Conclusion: 9 people are wearing caps.
Question 10
Laptops, Tablets & Books
In a training session, 17 participants have laptops, 11 have tablets, 9 have laptops and tablets, 6 have laptops and books, and 4 have both tablets and books. Eight participants have all three items. The total number of participants with laptops, tablets, or books is 35. How many participants have books?
Solution
Let L = laptops, T = tablets, and B = books.
Given values:
n(L) = 17, n(T) = 11, n(B) = x.
n(L ∩ T) = 9, n(L ∩ B) = 6, n(T ∩ B) = 4.
Triple overlap: n(L ∩ T ∩ B) = 8.
Total union: n(L ∪ T ∪ B) = 35.
n(L) = 17, n(T) = 11, n(B) = x.
n(L ∩ T) = 9, n(L ∩ B) = 6, n(T ∩ B) = 4.
Triple overlap: n(L ∩ T ∩ B) = 8.
Total union: n(L ∪ T ∪ B) = 35.
Apply Principle of Inclusion-Exclusion:
n(L ∪ T ∪ B) = n(L) + n(T) + n(B) - n(L ∩ T) - n(L ∩ B) - n(T ∩ B) + n(L ∩ T ∩ B)
n(L ∪ T ∪ B) = n(L) + n(T) + n(B) - n(L ∩ T) - n(L ∩ B) - n(T ∩ B) + n(L ∩ T ∩ B)
Substitute values:
35 = 17 + 11 + x - 9 - 6 - 4 + 8
35 = 36 - 19 + x
35 = 17 + x
x = 35 - 17 = 18
Conclusion: 18 participants have books.
Question 11
Mall Salaries partition
A shopping mall has 150 employees labeled 1 to 150 representing the universal set U.
- Set A: 40 employees labeled from 50 to 89.
- Set B: 50 employees labeled from 101 to 150.
- Set C: 60 employees labeled from 1 to 49 and 90 to 100.
Find: (a) (A' ∪ B') ∩ C and (b) n{A ∩ (B' ∩ C')}.
- Set A: 40 employees labeled from 50 to 89.
- Set B: 50 employees labeled from 101 to 150.
- Set C: 60 employees labeled from 1 to 49 and 90 to 100.
Find: (a) (A' ∪ B') ∩ C and (b) n{A ∩ (B' ∩ C')}.
Solution
Notice that Sets A, B, and C are pairwise disjoint and partition the universal set U:
A ∩ B = ∅, B ∩ C = ∅, A ∩ C = ∅, and A ∪ B ∪ C = U.
A ∩ B = ∅, B ∩ C = ∅, A ∩ C = ∅, and A ∪ B ∪ C = U.
(a) Find (A' ∪ B') ∩ C:
Since A, B, C partition U:
A' = B ∪ C and B' = A ∪ C.
Therefore, A' ∪ B' = (B ∪ C) ∪ (A ∪ C) = A ∪ B ∪ C = U.
Then, (A' ∪ B') ∩ C = U ∩ C = C.
So, (A' ∪ B') ∩ C = C = {1, 2, 3, ..., 49, 90, 91, ..., 100}.
Since A, B, C partition U:
A' = B ∪ C and B' = A ∪ C.
Therefore, A' ∪ B' = (B ∪ C) ∪ (A ∪ C) = A ∪ B ∪ C = U.
Then, (A' ∪ B') ∩ C = U ∩ C = C.
So, (A' ∪ B') ∩ C = C = {1, 2, 3, ..., 49, 90, 91, ..., 100}.
(b) Find n{A ∩ (B' ∩ C')}:
Compute B' ∩ C':
B' = A ∪ C and C' = A ∪ B.
Their intersection is: B' ∩ C' = (A ∪ C) ∩ (A ∪ B) = A.
Therefore, A ∩ (B' ∩ C') = A ∩ A = A.
The number of elements is: n(A) = 40.
Compute B' ∩ C':
B' = A ∪ C and C' = A ∪ B.
Their intersection is: B' ∩ C' = (A ∪ C) ∩ (A ∪ B) = A.
Therefore, A ∩ (B' ∩ C') = A ∩ A = A.
The number of elements is: n(A) = 40.
Question 12
Sports Venn diagram
In a secondary school, 125 students participate in at least one of the following sports: cricket, football, or hockey.
- 60 students play cricket.
- 70 students play football.
- 40 students play hockey.
- 25 students play cricket and football.
- 15 students play football and hockey.
- 10 students play cricket and hockey.
(a) How many students play all three sports?
(b) Draw a Venn diagram.
- 60 students play cricket.
- 70 students play football.
- 40 students play hockey.
- 25 students play cricket and football.
- 15 students play football and hockey.
- 10 students play cricket and hockey.
(a) How many students play all three sports?
(b) Draw a Venn diagram.
Solution
Let C, F, and H represent cricket, football, and hockey players.
Given data:
Total union n(C ∪ F ∪ H) = 125.
Individual sizes: n(C) = 60, n(F) = 70, n(H) = 40.
Intersections of two sets: n(C ∩ F) = 25, n(F ∩ H) = 15, n(C ∩ H) = 10.
Total union n(C ∪ F ∪ H) = 125.
Individual sizes: n(C) = 60, n(F) = 70, n(H) = 40.
Intersections of two sets: n(C ∩ F) = 25, n(F ∩ H) = 15, n(C ∩ H) = 10.
Apply Principle of Inclusion-Exclusion:
n(C ∪ F ∪ H) = n(C) + n(F) + n(H) - n(C ∩ F) - n(F ∩ H) - n(C ∩ H) + n(C ∩ F ∩ H)
n(C ∪ F ∪ H) = n(C) + n(F) + n(H) - n(C ∩ F) - n(F ∩ H) - n(C ∩ H) + n(C ∩ F ∩ H)
Substitute values:
125 = 60 + 70 + 40 - 25 - 15 - 10 + n(C ∩ F ∩ H)
125 = 170 - 50 + n(C ∩ F ∩ H)
125 = 120 + n(C ∩ F ∩ H)
n(C ∩ F ∩ H) = 125 - 120 = 5
Conclusion: 5 students play all three sports.
Question 13
Food favorites survey
A survey of 130 people regarding food preferences showed:
- 40 people like nihari.
- 65 people like biryani.
- 50 people like korma.
- 20 people liked nihari and biryani.
- 35 people liked biryani and korma.
- 27 people liked nihari and korma.
- 12 people liked all three foods.
Find: (a) How many like Nihari, Biryani, or Korma? (b) How many do not like any? (c) How many like only one food item?
- 40 people like nihari.
- 65 people like biryani.
- 50 people like korma.
- 20 people liked nihari and biryani.
- 35 people liked biryani and korma.
- 27 people liked nihari and korma.
- 12 people liked all three foods.
Find: (a) How many like Nihari, Biryani, or Korma? (b) How many do not like any? (c) How many like only one food item?
Solution
Let N = Nihari, B = Biryani, and K = Korma. Total survey n(U) = 130.
(a) Find n(N ∪ B ∪ K):
n(N ∪ B ∪ K) = n(N) + n(B) + n(K) - n(N ∩ B) - n(B ∩ K) - n(N ∩ K) + n(N ∩ B ∩ K)
= 40 + 65 + 50 - 20 - 35 - 27 + 12
= 167 - 82 = 85
So, 85 people like at least one of the foods.
(b) Find how many like none:
n(None) = n(U) - n(N ∪ B ∪ K) = 130 - 85 = 45.
n(None) = n(U) - n(N ∪ B ∪ K) = 130 - 85 = 45.
(c) Find how many like only one food item:
- Only Nihari: n(N) - n(N ∩ B) - n(N ∩ K) + n(N ∩ B ∩ K) = 40 - 20 - 27 + 12 = 5.
- Only Biryani: n(B) - n(N ∩ B) - n(B ∩ K) + n(N ∩ B ∩ K) = 65 - 20 - 35 + 12 = 22.
- Only Korma: n(K) - n(N ∩ K) - n(B ∩ K) + n(N ∩ B ∩ K) = 50 - 27 - 35 + 12 = 0.
Total who like only one item: 5 + 22 + 0 = 27.
- Only Nihari: n(N) - n(N ∩ B) - n(N ∩ K) + n(N ∩ B ∩ K) = 40 - 20 - 27 + 12 = 5.
- Only Biryani: n(B) - n(N ∩ B) - n(B ∩ K) + n(N ∩ B ∩ K) = 65 - 20 - 35 + 12 = 22.
- Only Korma: n(K) - n(N ∩ K) - n(B ∩ K) + n(N ∩ B ∩ K) = 50 - 27 - 35 + 12 = 0.
Total who like only one item: 5 + 22 + 0 = 27.
Interactive Sandbox
Explore Set Operations & Venn Diagrams
Configure the sizes of Set A, Set B, Set C, and their intersections below. The SVG Venn diagram and the Principle of Inclusion-Exclusion calculations will update in real-time. Toggle different operations to highlight their regions.
Principle of Inclusion-Exclusion Breakdown