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Logical Reasoning · Logical Connectives, Syllogisms and Venn Diagrams

Venn Diagram - Formula, Key Point and Examples

Explanation

🔵
VENN DIAGRAM\text{VENN DIAGRAM}
🟢 1. BASIC CONCEPT A Venn diagram is a graphical representation of sets using circles or closed curves.
Venn Diagram=Graphical representation of sets \text{Venn Diagram} = \text{Graphical representation of sets}
The important concepts are:
Union,Intersection,Difference,Complement \text{Union},\quad \text{Intersection},\quad \text{Difference},\quad \text{Complement}
🟢 2. SET A set is a collection of well-defined objects. For example:
A={1,2,3,4,5} A=\{1,2,3,4,5\}
The number of elements in set \(A\) is:
n(A)=5 n(A)=5
🟢 3. UNIVERSAL SET The universal set contains all the elements under consideration. It is represented by:
U U
🟢 4. UNION OF TWO SETS The union of sets \(A\) and \(B\) contains all elements belonging to \(A\), \(B\), or both. It is represented by:
A∪B A\cup B
The formula is:
n(A∪B)=n(A)+n(B)−n(A∩B) n(A\cup B) = n(A)+n(B)-n(A\cap B)
🟢 5. INTERSECTION OF TWO SETS The intersection contains the elements common to both sets. It is represented by:
A∩B A\cap B
🟢 6. DIFFERENCE OF SETS The difference \(A-B\) contains elements that belong to \(A\) but not to \(B\).
A−B A-B
Number of elements:
n(A−B)=n(A)−n(A∩B) n(A-B) = n(A)-n(A\cap B)
Similarly:
n(B−A)=n(B)−n(A∩B) n(B-A) = n(B)-n(A\cap B)
🟢 7. COMPLEMENT OF A SET The complement of \(A\) contains all elements of the universal set that are not in \(A\). It is represented by:
A′ A'
or:
Ac A^c
Formula:
n(A′)=n(U)−n(A) n(A') = n(U)-n(A)
🟢 8. TWO-SET VENN DIAGRAM For two sets \(A\) and \(B\), the important regions are:
A only A\text{ only}
B only B\text{ only}
A∩B A\cap B
Neither A nor B \text{Neither }A\text{ nor }B
🟢 9. ONLY A The elements belonging to \(A\) but not to \(B\) are:
A−B A-B
Therefore:
n(A only)=n(A)−n(A∩B) n(A\text{ only}) = n(A)-n(A\cap B)
🟢 10. ONLY B The elements belonging to \(B\) but not to \(A\) are:
B−A B-A
Therefore:
n(B only)=n(B)−n(A∩B) n(B\text{ only}) = n(B)-n(A\cap B)
🟢 11. NEITHER A NOR B Elements belonging to neither \(A\) nor \(B\) are outside the union.
n(Neither)=n(U)−n(A∪B) n(\text{Neither}) = n(U)-n(A\cup B)
Therefore:
n(Neither)=n(U)−n(A)−n(B)+n(A∩B) n(\text{Neither}) = n(U)-n(A)-n(B)+n(A\cap B)
🟢 12. UNION FORMULA For two sets:
n(A∪B)=n(A)+n(B)−n(A∩B) n(A\cup B) = n(A)+n(B)-n(A\cap B)
🔴 IMPORTANT The intersection is subtracted because common elements are counted twice.
Union=First Set+Second Set−Common Elements \text{Union} = \text{First Set} + \text{Second Set} - \text{Common Elements}
🟢 13. FINDING THE INTERSECTION From the union formula:
n(A∩B)=n(A)+n(B)−n(A∪B) n(A\cap B) = n(A)+n(B)-n(A\cup B)
🟢 14. DISJOINT SETS Two sets are disjoint if they have no common elements. Therefore:
A∩B=∅ A\cap B=\varnothing
and:
n(A∩B)=0 n(A\cap B)=0
Hence:
n(A∪B)=n(A)+n(B) n(A\cup B) = n(A)+n(B)
🟢 15. THREE-SET VENN DIAGRAM For three sets \(A\), \(B\), and \(C\), the important regions are:
A only A\text{ only}
B only B\text{ only}
C only C\text{ only}
A∩B only A\cap B\text{ only}
A∩C only A\cap C\text{ only}
B∩C only B\cap C\text{ only}
A∩B∩C A\cap B\cap C
and:
None \text{None}
🟢 16. THREE-SET UNION FORMULA For three sets:
n(A∪B∪C)=n(A)+n(B)+n(C) n(A\cup B\cup C) = n(A)+n(B)+n(C)
−n(A∩B)−n(A∩C)−n(B∩C) -n(A\cap B) -n(A\cap C) -n(B\cap C)
+n(A∩B∩C) +n(A\cap B\cap C)
🔴 IMPORTANT For three sets, the common intersection of all three sets is added once. 🟢 17. ONLY A IN THREE SETS The number belonging only to \(A\) is:
n(A only)=n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C) n(A\text{ only}) = n(A) -n(A\cap B) -n(A\cap C) +n(A\cap B\cap C)
🟢 18. ONLY B IN THREE SETS
n(B only)=n(B)−n(A∩B)−n(B∩C)+n(A∩B∩C) n(B\text{ only}) = n(B) -n(A\cap B) -n(B\cap C) +n(A\cap B\cap C)
🟢 19. ONLY C IN THREE SETS
n(C only)=n(C)−n(A∩C)−n(B∩C)+n(A∩B∩C) n(C\text{ only}) = n(C) -n(A\cap C) -n(B\cap C) +n(A\cap B\cap C)
🟢 20. A AND B ONLY Elements belonging to \(A\) and \(B\), but not \(C\):
n(A∩B only)=n(A∩B)−n(A∩B∩C) n(A\cap B\text{ only}) = n(A\cap B) -n(A\cap B\cap C)
🟢 21. A AND C ONLY
n(A∩C only)=n(A∩C)−n(A∩B∩C) n(A\cap C\text{ only}) = n(A\cap C) -n(A\cap B\cap C)
🟢 22. B AND C ONLY
n(B∩C only)=n(B∩C)−n(A∩B∩C) n(B\cap C\text{ only}) = n(B\cap C) -n(A\cap B\cap C)
🟢 23. ALL THREE SETS Elements belonging to all three sets are represented by:
A∩B∩C A\cap B\cap C
Therefore:
n(All Three)=n(A∩B∩C) n(\text{All Three}) = n(A\cap B\cap C)
🟢 24. NONE OF THE THREE SETS The number of elements belonging to none of the three sets is:
n(None)=n(U)−n(A∪B∪C) n(\text{None}) = n(U)-n(A\cup B\cup C)
🟢 25. AT LEAST ONE "At least one" means belonging to one or more sets. Therefore:
At Least One=A∪B∪C \text{At Least One} = A\cup B\cup C
Hence:
n(At Least One)=n(A∪B∪C) n(\text{At Least One}) = n(A\cup B\cup C)
🟢 26. AT LEAST TWO "At least two" means belonging to two or three sets.
n(At Least Two)=n(A∩B)+n(A∩C)+n(B∩C) n(\text{At Least Two}) = n(A\cap B) +n(A\cap C) +n(B\cap C)
Since the elements belonging to all three sets are counted three times:
n(At Least Two)=n(A∩B)+n(A∩C)+n(B∩C)−2n(A∩B∩C) n(\text{At Least Two}) = n(A\cap B) +n(A\cap C) +n(B\cap C) -2n(A\cap B\cap C)
🟢 27. EXACTLY TWO "Exactly two" means belonging to exactly two sets but not all three.
n(Exactly Two)=n(A∩B)+n(A∩C)+n(B∩C)−3n(A∩B∩C) n(\text{Exactly Two}) = n(A\cap B) +n(A\cap C) +n(B\cap C) -3n(A\cap B\cap C)
🟢 28. EXACTLY ONE "Exactly one" means belonging to only one of the three sets.
n(Exactly One)=n(A only)+n(B only)+n(C only) n(\text{Exactly One}) = n(A\text{ only}) +n(B\text{ only}) +n(C\text{ only})
🟢 29. AT LEAST ONE AND NONE For the universal set:
n(At Least One)+n(None)=n(U) n(\text{At Least One}) + n(\text{None}) = n(U)
Therefore:
n(None)=n(U)−n(At Least One) n(\text{None}) = n(U)-n(\text{At Least One})
🟢 30. IMPORTANT VENN DIAGRAM SYMBOLS
∪=Union \cup = \text{Union}
∩=Intersection \cap = \text{Intersection}
A−B=Difference A-B = \text{Difference}
A′=Complement of A A' = \text{Complement of A}
∅=Empty Set \varnothing = \text{Empty Set}
U=Universal Set U = \text{Universal Set}
🔴 IMPORTANT KEY POINTS
A∪B=Elements in A or B or both A\cup B = \text{Elements in A or B or both}
A∩B=Elements common to A and B A\cap B = \text{Elements common to A and B}
A−B=Elements in A but not in B A-B = \text{Elements in A but not in B}
A′=Elements not in A A' = \text{Elements not in A}
n(A∪B)=n(A)+n(B)−n(A∩B) n(A\cup B) = n(A)+n(B)-n(A\cap B)
n(A only)=n(A)−n(A∩B) n(A\text{ only}) = n(A)-n(A\cap B)
n(B only)=n(B)−n(A∩B) n(B\text{ only}) = n(B)-n(A\cap B)
n(Neither)=n(U)−n(A∪B) n(\text{Neither}) = n(U)-n(A\cup B)
n(A∪B∪C)=n(A)+n(B)+n(C) n(A\cup B\cup C) = n(A)+n(B)+n(C)
−n(A∩B)−n(A∩C)−n(B∩C) -n(A\cap B) -n(A\cap C) -n(B\cap C)
+n(A∩B∩C) +n(A\cap B\cap C)
At Least One=A∪B∪C \text{At Least One} = A\cup B\cup C
None=U−(A∪B∪C) \text{None} = U-(A\cup B\cup C)
Exactly Two=Pairwise Intersections−3×Triple Intersection \text{Exactly Two} = \text{Pairwise Intersections} - 3\times\text{Triple Intersection}
Always subtract overlapping regions when calculating a union \text{Always subtract overlapping regions when calculating a union}
For three sets, add the triple intersection once \text{For three sets, add the triple intersection once}

Example

🔵
VENN DIAGRAM — SOLVED EXAMPLES\text{VENN DIAGRAM — SOLVED EXAMPLES}
🟢 EXAMPLE 1: UNION OF TWO SETS QUESTION: In a class of 50 students, 30 students like Mathematics and 25 students like Science. If 10 students like both subjects, how many students like at least one of the two subjects? SOLUTION:
n(A∪B)=n(A)+n(B)−n(A∩B) n(A\cup B) = n(A)+n(B)-n(A\cap B)
=30+25−10 = 30+25-10
=45 =45
🔵 ANSWER:
45 students 45\text{ students}
🟢 EXAMPLE 2: STUDENTS WHO LIKE NEITHER SUBJECT QUESTION: In a class of 60 students, 35 like Mathematics, 30 like Science, and 15 like both. How many students like neither subject? SOLUTION:
n(A∪B)=n(A)+n(B)−n(A∩B) n(A\cup B) = n(A)+n(B)-n(A\cap B)
=35+30−15 = 35+30-15
=50 =50
Therefore:
n(Neither)=n(U)−n(A∪B) n(\text{Neither}) = n(U)-n(A\cup B)
=60−50 =60-50
=10 =10
🔵 ANSWER:
10 students 10\text{ students}
🟢 EXAMPLE 3: ONLY MATHEMATICS QUESTION: In a group of 80 students, 45 like Mathematics, 30 like English, and 12 like both. How many students like only Mathematics? SOLUTION:
n(A only)=n(A)−n(A∩B) n(A\text{ only}) = n(A)-n(A\cap B)
=45−12 =45-12
=33 =33
🔵 ANSWER:
33 students 33\text{ students}
🟢 EXAMPLE 4: ONLY ENGLISH QUESTION: In a group of 70 students, 40 like English, 25 like Hindi, and 10 like both. How many students like only Hindi? SOLUTION:
n(B only)=n(B)−n(A∩B) n(B\text{ only}) = n(B)-n(A\cap B)
=25−10 =25-10
=15 =15
🔵 ANSWER:
15 students 15\text{ students}
🟢 EXAMPLE 5: FINDING THE INTERSECTION QUESTION: In a group of 100 students, 60 like Cricket, 50 like Football, and 80 like at least one of the two sports. How many students like both Cricket and Football? SOLUTION:
n(A∪B)=n(A)+n(B)−n(A∩B) n(A\cup B) = n(A)+n(B)-n(A\cap B)
Therefore:
n(A∩B)=n(A)+n(B)−n(A∪B) n(A\cap B) = n(A)+n(B)-n(A\cup B)
=60+50−80 =60+50-80
=30 =30
🔵 ANSWER:
30 students 30\text{ students}
🟢 EXAMPLE 6: NEITHER OF TWO SETS QUESTION: Out of 120 people, 70 read Newspaper A, 60 read Newspaper B, and 40 read both. How many people read neither newspaper? SOLUTION:
n(A∪B)=n(A)+n(B)−n(A∩B) n(A\cup B) = n(A)+n(B)-n(A\cap B)
=70+60−40 =70+60-40
=90 =90
Therefore:
n(Neither)=120−90 n(\text{Neither}) = 120-90
=30 =30
🔵 ANSWER:
30 people 30\text{ people}
🟢 EXAMPLE 7: THREE SETS QUESTION: In a group of 100 students, 50 study Mathematics, 45 study Physics, and 40 study Chemistry. 20 study both Mathematics and Physics, 15 study both Physics and Chemistry, 18 study both Mathematics and Chemistry, and 8 study all three subjects. How many students study at least one subject? SOLUTION:
n(A∪B∪C)=n(A)+n(B)+n(C) n(A\cup B\cup C) = n(A)+n(B)+n(C)
−n(A∩B)−n(A∩C)−n(B∩C) -n(A\cap B) -n(A\cap C) -n(B\cap C)
+n(A∩B∩C) +n(A\cap B\cap C)
Substituting:
=50+45+40−20−18−15+8 =50+45+40-20-18-15+8
=90 =90
🔵 ANSWER:
90 students 90\text{ students}
🟢 EXAMPLE 8: NONE OF THREE SETS QUESTION: In a group of 100 students, 50 study Mathematics, 45 study Physics, and 40 study Chemistry. 20 study Mathematics and Physics, 15 study Physics and Chemistry, 18 study Mathematics and Chemistry, and 8 study all three. How many students study none of these subjects? SOLUTION: From the previous calculation:
n(A∪B∪C)=90 n(A\cup B\cup C)=90
Therefore:
n(None)=n(U)−n(A∪B∪C) n(\text{None}) = n(U)-n(A\cup B\cup C)
=100−90 =100-90
=10 =10
🔵 ANSWER:
10 students 10\text{ students}
🟢 EXAMPLE 9: ONLY ONE SET QUESTION: In a group of 100 students:
n(A)=50 n(A)=50
n(B)=45 n(B)=45
n(C)=40 n(C)=40
n(A∩B)=20 n(A\cap B)=20
n(A∩C)=18 n(A\cap C)=18
n(B∩C)=15 n(B\cap C)=15
n(A∩B∩C)=8 n(A\cap B\cap C)=8
Find the number of students who study only Mathematics. SOLUTION:
n(A only)=n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C) n(A\text{ only}) = n(A) -n(A\cap B) -n(A\cap C) +n(A\cap B\cap C)
=50−20−18+8 =50-20-18+8
=20 =20
🔵 ANSWER:
20 students 20\text{ students}
🟢 EXAMPLE 10: ONLY PHYSICS QUESTION: Using the same data, find the number of students who study only Physics. SOLUTION:
n(B only)=n(B)−n(A∩B)−n(B∩C)+n(A∩B∩C) n(B\text{ only}) = n(B) -n(A\cap B) -n(B\cap C) +n(A\cap B\cap C)
=45−20−15+8 =45-20-15+8
=18 =18
🔵 ANSWER:
18 students 18\text{ students}
🟢 EXAMPLE 11: ONLY CHEMISTRY QUESTION: Using the same data, find the number of students who study only Chemistry. SOLUTION:
n(C only)=n(C)−n(A∩C)−n(B∩C)+n(A∩B∩C) n(C\text{ only}) = n(C) -n(A\cap C) -n(B\cap C) +n(A\cap B\cap C)
=40−18−15+8 =40-18-15+8
=15 =15
🔵 ANSWER:
15 students 15\text{ students}
🟢 EXAMPLE 12: MATHEMATICS AND PHYSICS ONLY QUESTION: In a group, 25 students study both Mathematics and Physics, and 10 students study all three subjects. How many study Mathematics and Physics but not Chemistry? SOLUTION:
n(A∩B only)=n(A∩B)−n(A∩B∩C) n(A\cap B\text{ only}) = n(A\cap B)-n(A\cap B\cap C)
=25−10 =25-10
=15 =15
🔵 ANSWER:
15 students 15\text{ students}
🟢 EXAMPLE 13: PHYSICS AND CHEMISTRY ONLY QUESTION: 30 students study both Physics and Chemistry, while 12 students study all three subjects. How many study Physics and Chemistry but not Mathematics? SOLUTION:
n(B∩C only)=n(B∩C)−n(A∩B∩C) n(B\cap C\text{ only}) = n(B\cap C)-n(A\cap B\cap C)
=30−12 =30-12
=18 =18
🔵 ANSWER:
18 students 18\text{ students}
🟢 EXAMPLE 14: ALL THREE SETS QUESTION: In a survey, 35 people like Tea, 30 like Coffee, and 25 like Juice. If 12 people like all three, how many people like all three beverages? SOLUTION: The number of people who like all three is directly given:
n(A∩B∩C)=12 n(A\cap B\cap C)=12
🔵 ANSWER:
12 people 12\text{ people}
🟢 EXAMPLE 15: AT LEAST ONE QUESTION: In a group, 40 students play Cricket, 35 play Football, 30 play Basketball, 15 play Cricket and Football, 12 play Football and Basketball, 10 play Cricket and Basketball, and 5 play all three. Find the number of students who play at least one game. SOLUTION:
n(A∪B∪C)=n(A)+n(B)+n(C) n(A\cup B\cup C) = n(A)+n(B)+n(C)
−n(A∩B)−n(A∩C)−n(B∩C) -n(A\cap B) -n(A\cap C) -n(B\cap C)
+n(A∩B∩C) +n(A\cap B\cap C)
=40+35+30−15−10−12+5 =40+35+30-15-10-12+5
=73 =73
🔵 ANSWER:
73 students 73\text{ students}
🟢 EXAMPLE 16: EXACTLY TWO QUESTION: In a group, 20 students study both Mathematics and Physics, 18 study both Physics and Chemistry, 15 study both Mathematics and Chemistry, and 5 study all three subjects. How many students study exactly two subjects? SOLUTION:
n(Exactly Two)=n(A∩B)+n(A∩C)+n(B∩C) n(\text{Exactly Two}) = n(A\cap B) +n(A\cap C) +n(B\cap C)
−3n(A∩B∩C) -3n(A\cap B\cap C)
=20+15+18−3(5) =20+15+18-3(5)
=53−15 =53-15
=38 =38
🔵 ANSWER:
38 students 38\text{ students}
🟢 EXAMPLE 17: AT LEAST TWO QUESTION: In a group, 25 students like Mathematics and Physics, 20 like Mathematics and Chemistry, 18 like Physics and Chemistry, and 6 like all three. How many students like at least two subjects? SOLUTION:
n(At Least Two)=n(A∩B)+n(A∩C)+n(B∩C) n(\text{At Least Two}) = n(A\cap B) +n(A\cap C) +n(B\cap C)
−2n(A∩B∩C) -2n(A\cap B\cap C)
=25+20+18−2(6) =25+20+18-2(6)
=63−12 =63-12
=51 =51
🔵 ANSWER:
51 students 51\text{ students}
🟢 EXAMPLE 18: EXACTLY ONE QUESTION: In a group of students:
n(A only)=20 n(A\text{ only})=20
n(B only)=15 n(B\text{ only})=15
n(C only)=10 n(C\text{ only})=10
Find the number of students who study exactly one subject. SOLUTION:
n(Exactly One)=n(A only)+n(B only)+n(C only) n(\text{Exactly One}) = n(A\text{ only}) +n(B\text{ only}) +n(C\text{ only})
=20+15+10 =20+15+10
=45 =45
🔵 ANSWER:
45 students 45\text{ students}
🟢 EXAMPLE 19: DISJOINT SETS QUESTION: In a class, 25 students like Mathematics and 20 students like History. No student likes both subjects. Find the number of students who like at least one subject. SOLUTION: Since no student likes both:
A∩B=∅ A\cap B=\varnothing
Therefore:
n(A∩B)=0 n(A\cap B)=0
Using:
n(A∪B)=n(A)+n(B)−n(A∩B) n(A\cup B) = n(A)+n(B)-n(A\cap B)
=25+20−0 =25+20-0
=45 =45
🔵 ANSWER:
45 students 45\text{ students}
🟢 EXAMPLE 20: MIXED VENN DIAGRAM PROBLEM QUESTION: In a group of 150 students, 80 like Cricket, 70 like Football, and 60 like Basketball. 30 like Cricket and Football, 25 like Football and Basketball, 20 like Cricket and Basketball, and 10 like all three. Find the number of students who like none of the three games. SOLUTION: First find the number who like at least one game.
n(A∪B∪C)=n(A)+n(B)+n(C) n(A\cup B\cup C) = n(A)+n(B)+n(C)
−n(A∩B)−n(A∩C)−n(B∩C) -n(A\cap B) -n(A\cap C) -n(B\cap C)
+n(A∩B∩C) +n(A\cap B\cap C)
=80+70+60−30−20−25+10 =80+70+60-30-20-25+10
=145 =145
Therefore:
n(None)=n(U)−n(A∪B∪C) n(\text{None}) = n(U)-n(A\cup B\cup C)
=150−145 =150-145
=5 =5
🔵 ANSWER:
5 students 5\text{ students}