Principle of Inclusion and Exclusion (PIE)
The Principle of Inclusion and Exclusion (PIE) is a fundamental counting principle in mathematics, combinatorics, and set theory. It helps in calculating the number of elements in the union of several sets by adding the sizes of the sets and subtracting their overlaps, so that no element is counted more than once.
Definition
If we have two or more sets, the principle of inclusion and exclusion gives a systematic way to compute the size of their union. The method alternates between inclusion (adding sizes of sets) and exclusion (subtracting intersections) until all overlaps are correctly handled.
Formulas
For Two Sets:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
For Three Sets:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − [n(A ∩ B) + n(B ∩ C) + n(C ∩ A)] + n(A ∩ B ∩ C)
General Case (n sets):
n(A1 ∪ A2 ∪ ... ∪ An) = Σ n(Ai) − Σ n(Ai ∩ Aj) + Σ n(Ai ∩ Aj ∩ Ak) − ... + (−1)n+1 n(A1 ∩ A2 ∩ ... ∩ An)
Applications of PIE
- Survey problems – finding how many people like at least one subject.
- Probability problems – computing the probability that at least one event occurs.
- Divisibility problems – counting integers divisible by given primes.
- Data verification – detecting inconsistencies in survey or census data.
Conclusion
The Principle of Inclusion and Exclusion is an essential method in set theory, combinatorics, and probability. It prevents overcounting by systematically adjusting for overlaps. By learning PIE and the formulas for “only” parts, students can solve a wide range of problems in both school exams and competitive exams with confidence.
Formulas for “Only” Parts
For problems involving exclusive regions, the following are useful:
• Only A (not B, not C) = n(A) − n(A ∩ B) − n(A ∩ C) + n(A ∩ B ∩ C)
• Only B (not A, not C) = n(B) − n(A ∩ B) − n(B ∩ C) + n(A ∩ B ∩ C)
• Only C (not A, not B) = n(C) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C)
• A ∩ B only (not C) = n(A ∩ B) − n(A ∩ B ∩ C)
• B ∩ C only (not A) = n(B ∩ C) − n(A ∩ B ∩ C)
• C ∩ A only (not B) = n(C ∩ A) − n(A ∩ B ∩ C)
• All three (A ∩ B ∩ C) = n(A ∩ B ∩ C)
Illustrative Examples on Sets
Q1). If S has 21 elements, T has 32 elements and S ∩ T has 11 elements, find n(S ∪ T).
Solution:
⇒ 21 + 32 – 11
⇒ 42
Answer: n(S ∪ T) = 42
Q2). Suppose n(X ∪ Y) = 18, n(X) = 12, and n(X ∩ Y) = 5. Find n(Y).
Solution:
⇒ n(Y) = n(X ∪ Y) – n(X) + n(X ∩ Y)
⇒ 18 – 12 + 5
⇒ 11
Answer: n(Y) = 11
Q3). If n(A) = 40, n(B) = 30, and n(A ∩ B) = 10, find n(A ∪ B) and n(A ∖ B).
Solution:
⇒ 40 + 30 – 10 = 60
n(A ∖ B) = n(A) – n(A ∩ B)
⇒ 40 – 10 = 30
Answer: n(A ∪ B) = 60, n(A ∖ B) = 30
Q4). Let n(A) = 17, n(B) = 23, n(C) = 12, n(A ∩ B) = 5, n(B ∩ C) = 3, n(C ∩ A) = 4, n(A ∩ B ∩ C) = 2. Find n(A ∪ B ∪ C).
Solution:
– n(A ∩ B) – n(B ∩ C) – n(C ∩ A)
+ n(A ∩ B ∩ C)
= 17 + 23 + 12 – (5 + 3 + 4) + 2
⇒ 52 – 12 + 2
⇒ 42
Answer: n(A ∪ B ∪ C) = 42
Q5). Let A and B be two sets such that n(A) = 40, n(B) = 62 and n(A ∪ B) = 95. Find n(A ∩ B), n(A – B), n(B – A), and n(A Δ B).
Solution:
⇒ 95 = 40 + 62 – n(A ∩ B)
⇒ n(A ∩ B) = 102 – 95 = 7
n(A – B) = n(A) – n(A ∩ B) = 40 – 7 = 33
n(B – A) = n(B) – n(A ∩ B) = 62 – 7 = 55
n(A Δ B) = n(A – B) + n(B – A) = 33 + 55 = 88
Answer: n(A ∩ B) = 7, n(A – B) = 33, n(B – A) = 55, n(A Δ B) = 88
Q6). In a group of 500 people, 280 can speak Hindi and 260 can speak English. How many can speak both Hindi and English?
Solution:
n(H) = 280, n(E) = 260, n(H ∪ E) = 500
n(H ∩ E) = n(H) + n(E) – n(H ∪ E)
= 280 + 260 – 500 = 40
Answer: 40 people can speak both Hindi and English.
Q7). Out of 80 persons in a class, 30 like cricket, 28 like football and 12 like both. Find how many like at least one of the two games.
Solution:
n(C ∪ F) = n(C) + n(F) – n(C ∩ F)
= 30 + 28 – 12 = 46
Answer: 46 students like at least one game.
Q8). If n(A) = 150, n(B) = 220 and n(U) = 600, find n(A′ ∩ B′).
Solution:
n(A′ ∩ B′) = n((A ∪ B)′) = n(U) – n(A ∪ B)
= n(U) – [n(A) + n(B) – n(A ∩ B)]
If A and B are disjoint (worst case):
n(A ∪ B) = 150 + 220 = 370
⇒ n(A′ ∩ B′) = 600 – 370 = 230
Answer: n(A′ ∩ B′) = 230
Q9). Out of 600 car owners investigated, 420 owned car A and 250 owned car B, 80 owned both. Is this data correct?
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 420 + 250 – 80 = 590
Since n(A ∪ B) ≤ n(U) and 590 ≤ 600, the data is possible.
Answer: The data is correct.
Q10). In a group of 60 people, 40 speak Hindi, 28 speak English, and 18 speak both Hindi and English. How many people speak only English? How many people speak at least one language?
Solution:
We are given: n(U) = 60, n(H) = 40, n(E) = 28, n(H ∩ E) = 18.
People who speak only English = n(E – H)
= n(E) – n(H ∩ E)
= 28 – 18 = 10
People who speak at least one language = n(H ∪ E)
= n(H) + n(E) – n(H ∩ E)
= 40 + 28 – 18 = 50
Answer: 10 people speak only English, and 50 people speak at least one language.
Q11). There are 45 students in a Chemistry class and 65 students in a Physics class. Find the number of students who are in at least one of these classes in the following cases: (i) No student takes both classes. (ii) 15 students take both Chemistry and Physics.
Solution:
n(A) = 45, n(B) = 65.
(i) If no student takes both subjects, n(A ∩ B) = 0.
Then n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 45 + 65 – 0 = 110
→ All 110 are in at least one class.
(ii) If 15 students take both subjects, n(A ∩ B) = 15.
Then n(A ∪ B) = 45 + 65 – 15 = 95
→ Here 95 students are in at least one class.
Answer: (i) 110 students, (ii) 95 students.
Q12). In a survey of 800 students, 200 were drinking Limca, 320 were drinking Miranda, and 120 were drinking both Limca and Miranda. Find how many students were drinking neither of the two drinks.
Solution:
n(U) = 800, n(A) = 200, n(B) = 320, n(A ∩ B) = 120.
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 200 + 320 – 120 = 400
Students drinking neither = n(U) – n(A ∪ B)
= 800 – 400 = 400
Answer: 400 students drank neither Limca nor Miranda.
Q13). There are 220 individuals with a skin disorder. 130 have been exposed to chemical C1, 70 to chemical C2, and 40 to both chemicals. Find the number of individuals: (i) exposed to C1 or C2, (ii) exposed to C1 but not C2, (iii) exposed to C2 but not C1.
Solution:
n(U) = 220, n(A) = 130, n(B) = 70, n(A ∩ B) = 40.
(i) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 130 + 70 – 40 = 160
⇒ 160 individuals exposed to C1 or C2.
(ii) Exposed to C1 but not C2 = n(A – B)
= n(A) – n(A ∩ B)
= 130 – 40 = 90
(iii) Exposed to C2 but not C1 = n(B – A)
= n(B) – n(A ∩ B)
= 70 – 40 = 30
Answer: (i) 160 individuals, (ii) 90 individuals, (iii) 30 individuals.
Q14). Out of 90 students, 55 have taken Mathematics, 38 have taken Economics, and 22 have taken both Mathematics and Economics. Find the number of students who have taken: (i) Mathematics but not Economics, (ii) Economics but not Mathematics, (iii) Either Mathematics or Economics but not both.
Solution:
(i) Mathematics but not Economics = n(M – E) = n(M) – n(M ∩ E)
= 55 – 22 = 33
(ii) Economics but not Mathematics = n(E – M) = n(E) – n(M ∩ E)
= 38 – 22 = 16
(iii) Either Mathematics or Economics but not both = n(M Δ E)
= n(M – E) + n(E – M) = 33 + 16 = 49
Answer: (i) 33, (ii) 16, (iii) 49
Q15). If A and B are finite sets such that n(A) = 25 and n(B) = 20, find the least and greatest possible values of n(A ∪ B).
Solution:
- The greatest value of n(A ∪ B) occurs when A ∩ B = ∅ (disjoint).
⇒ n(A ∪ B) = n(A) + n(B) = 25 + 20 = 45
- The least value of n(A ∪ B) occurs when A ⊆ B or B ⊆ A.
⇒ n(A ∪ B) = max{n(A), n(B)} = max{25, 20} = 25
Answer: Least value = 25, Greatest value = 45
Q16). If A and B are two sets such that n(A) = 40, n(B) = 35, and n(U) = 80, find: (i) the least value of n(A ∪ B), (ii) the greatest value of n(A ∪ B).
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B).
Maximum overlap = min{n(A), n(B)} = 35.
⇒ n(A ∪ B) = 40 + 35 – 35 = 40.
(ii) Greatest value occurs when A ∩ B = ∅.
⇒ n(A ∪ B) = n(A) + n(B) = 40 + 35 = 75.
But n(A ∪ B) ≤ n(U) = 80, so valid.
Answer: Least value = 40, Greatest value = 75
Q17). If A and B are two sets such that n(A) = 28, n(B) = 36, and n(A ∩ B) = 14, find n(A ∪ B).
Solution:
⇒ n(A ∪ B) = 28 + 36 − 14
⇒ n(A ∪ B) = 50
Answer: n(A ∪ B) = 50
Q18). If P and Q are two sets such that P has 6 elements and P ∪ Q has 10 elements, how many elements does Q have if P ∩ Q has 4 elements?
Solution:
⇒ 10 = 6 + n(Q) − 4
⇒ 10 = 2 + n(Q)
⇒ n(Q) = 8
Answer: Q has 8 elements.
Q19). In a school there are 24 teachers. 12 teach Mathematics, 10 teach Physics, and 8 teach both Mathematics and Physics. How many teachers teach at least one of these subjects?
Solution:
n(M) = 12, n(P) = 10, n(M ∩ P) = 8.
n(M ∪ P) = n(M) + n(P) − n(M ∩ P)
⇒ 12 + 10 − 8 = 14
Answer: 14 teachers teach at least one of the two subjects.
Q20). In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the drinks. How many like both coffee and tea?
Solution:
Formula: n(Coffee ∪ Tea) = n(Coffee) + n(Tea) − n(Coffee ∩ Tea)
Since each person likes at least one drink ⇒ n(Coffee ∪ Tea) = n(U) = 70
⇒ 70 = 37 + 52 − n(Coffee ∩ Tea)
⇒ 70 = 89 − n(Coffee ∩ Tea)
⇒ n(Coffee ∩ Tea) = 89 − 70 = 19
Answer: 19 people like both coffee and tea.
Q21). Let A and B be two sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 42. Find: (i) n(A ∩ B) (ii) n(A − B) (iii) n(B − A)
Solution:
⇒ 42 = 20 + 28 − n(A ∩ B)
⇒ n(A ∩ B) = 48 − 42 = 6
(ii) n(A − B) = n(A) − n(A ∩ B)
⇒ 20 − 6 = 14
(iii) n(B − A) = n(B) − n(A ∩ B)
⇒ 28 − 6 = 22
Answer: n(A ∩ B) = 6, n(A − B) = 14, n(B − A) = 22
Q22). A survey shows that 63% of people like cold drinks and 76% like hot drinks. If each person likes at least one of the two drinks, find the percentage of people who like both cold and hot drinks.
Solution:
n(Cold) = 63, n(Hot) = 76.
Formula: n(Cold ∪ Hot) = n(Cold) + n(Hot) − n(Cold ∩ Hot)
But n(Cold ∪ Hot) = 100 (since every person likes at least one).
⇒ 100 = 63 + 76 − n(Cold ∩ Hot)
⇒ 100 = 139 − n(Cold ∩ Hot)
⇒ n(Cold ∩ Hot) = 39
Answer: 39% people like both cold and hot drinks.
Q23). In a group of 50 persons, 35 can speak Hindi and 30 can speak English. Find the number of persons who can speak: (i) Hindi only (ii) English only (iii) Both Hindi and English (iv) At least one of the two languages
Solution:
By formula: n(H ∪ E) = n(H) + n(E) − n(H ∩ E) ≤ n(U).
⇒ 35 + 30 − n(H ∩ E) ≤ 50
⇒ 65 − n(H ∩ E) ≤ 50
⇒ n(H ∩ E) ≥ 15.
Let’s assume minimum overlap 15 (so numbers fit correctly).
(i) Hindi only = n(H) − n(H ∩ E) = 35 − 15 = 20
(ii) English only = n(E) − n(H ∩ E) = 30 − 15 = 15
(iii) Both = 15
(iv) At least one = 20 + 15 + 15 = 50
Answer: Hindi only = 20, English only = 15, Both = 15, At least one = 50
Q24). In a group of 65 students, 20 play cricket, 25 play hockey, 15 play football, 10 play both cricket and hockey, 9 play both hockey and football, 4 play both football and cricket and 3 play all the three games. Find the number of students who play at least one of the three games.
Solution:
n(C ∩ H) = 10, n(H ∩ F) = 9, n(F ∩ C) = 4
n(C ∩ H ∩ F) = 3
Formula: n(C ∪ H ∪ F) = n(C) + n(H) + n(F)
− [n(C ∩ H) + n(H ∩ F) + n(F ∩ C)]
+ n(C ∩ H ∩ F)
⇒ 20 + 25 + 15 − (10 + 9 + 4) + 3
= 60 − 23 + 3
= 40
Answer: 40 students play at least one of the three games.
Q25). In a survey of 400 students, 100 were listed as taking apple juice, 150 as taking orange juice and 75 as taking both apple and orange juices. Find how many students were taking: (i) apple juice only (ii) orange juice only (iii) none of the two juices
Solution:
(i) Apple only = n(A) − n(A ∩ O) = 100 − 75 = 25
(ii) Orange only = n(O) − n(A ∩ O) = 150 − 75 = 75
(iii) At least one = n(A ∪ O) = n(A) + n(O) − n(A ∩ O)
= 100 + 150 − 75 = 175
None = n(U) − n(A ∪ O) = 400 − 175 = 225
Answer: Apple only = 25, Orange only = 75, None = 225
Q26). In a class of 60 students, 20 speak French, 20 speak Spanish and 10 speak both French and Spanish. Find how many students speak at least one of these two languages. Draw the Venn diagram.
Solution:
n(F ∪ S) = n(F) + n(S) − n(F ∩ S)
= 20 + 20 − 10
= 30
So, 30 students speak at least one language.
Answer: 30 students speak at least one language.
Q27). In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 as taking both apple as well as orange juice. Find how many students were taking: (i) apple juice only (ii) orange juice only (iii) none of the two juices
Solution:
(i) Apple only = n(A) − n(A ∩ O) = 100 − 75 = 25
(ii) Orange only = n(O) − n(A ∩ O) = 150 − 75 = 75
(iii) n(A ∪ O) = n(A) + n(O) − n(A ∩ O)
= 100 + 150 − 75 = 175
None = n(U) − n(A ∪ O) = 400 − 175 = 225
Answer: Apple only = 25, Orange only = 75, None = 225
Q28). In a class of 60 students, 20 speak French, 20 speak Spanish and 10 speak both Spanish and French. How many students speak at least one of these two languages?
Solution:
n(F ∪ S) = n(F) + n(S) − n(F ∩ S)
= 20 + 20 − 10
= 30
Answer: 30 students speak at least one of the two languages.
Q29). In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket, if 20 like tennis?
Solution:
Tennis only = n(Tennis) − n(C ∩ T)
= 20 − 10 = 10
Answer: 10 people like tennis only and not cricket.
Q30). In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 11 read both H and I, 9 read both H and T, 8 read both T and I and 3 read all the three newspapers. Find the number of people who read at least one newspaper.
Solution:
n(H) = 25, n(T) = 26, n(I) = 26
n(H ∩ I) = 11, n(H ∩ T) = 9, n(T ∩ I) = 8
n(H ∩ T ∩ I) = 3
By formula:
n(H ∪ T ∪ I) = n(H) + n(T) + n(I)
− [n(H ∩ T) + n(T ∩ I) + n(I ∩ H)]
+ n(H ∩ T ∩ I)
⇒ 25 + 26 + 26 − (9 + 8 + 11) + 3
= 77 − 28 + 3
= 52
Answer: 52 people read at least one newspaper.
Q31). Of the members of three athletic teams in a certain school, 21 are in the basketball team, 26 in the hockey team and 29 in the football team. 14 play hockey and basketball, 15 play hockey and football, 12 play football and basketball, and 8 play all the three games. How many students are there in all?
Solution:
n(H ∩ B) = 14, n(H ∩ F) = 15, n(F ∩ B) = 12
n(H ∩ F ∩ B) = 8
Formula: n(B ∪ H ∪ F) = n(B) + n(H) + n(F) − [n(B ∩ H) + n(H ∩ F) + n(F ∩ B)] + n(B ∩ H ∩ F)
⇒ 21 + 26 + 29 − (14 + 15 + 12) + 8
= 76 − 41 + 8
= 43
Answer: 43 students are there in all.
Q32). A survey of 500 television viewers produced the following information: 285 watch football, 195 watch hockey, 115 watch basketball, 50 watch football and basketball, 45 watch football and hockey, 50 watch hockey and basketball and 10 watch all the three games. How many watch at least one of the three games?
Solution:
n(F ∩ B) = 50, n(F ∩ H) = 45, n(H ∩ B) = 50
n(F ∩ H ∩ B) = 10
Formula: n(F ∪ H ∪ B) = n(F) + n(H) + n(B) − [n(F ∩ H) + n(H ∩ B) + n(F ∩ B)] + n(F ∩ H ∩ B)
⇒ 285 + 195 + 115 − (45 + 50 + 50) + 10
= 595 − 145 + 10
= 460
Answer: 460 viewers watch at least one of the three games.
Q33). In a survey of 108 persons, it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 10 read both A and B, 9 read both B and C, 8 read both C and A, and 5 read all the three magazines. Find the number of persons who read: (i) only one magazine (ii) exactly two magazines (iii) none of the magazines
Solution:
n(A) = 28, n(B) = 30, n(C) = 42
n(A ∩ B) = 10, n(B ∩ C) = 9, n(C ∩ A) = 8
n(A ∩ B ∩ C) = 5
Step 1: Exactly two-magazine counts:
(A ∩ B only) = n(A∩B) − n(A∩B∩C) = 10 − 5 = 5
(B ∩ C only) = n(B∩C) − n(A∩B∩C) = 9 − 5 = 4
(C ∩ A only) = n(C∩A) − n(A∩B∩C) = 8 − 5 = 3
Total with exactly two = 5 + 4 + 3 = 12
Step 2: Only one-magazine counts (using formula):
n(A only) = n(A) − n(A ∩ B) − n(A ∩ C) + n(A ∩ B ∩ C)
= 28 − 10 − 8 + 5 = 15
n(B only) = n(B) − n(A ∩ B) − n(B ∩ C) + n(A ∩ B ∩ C)
= 30 − 10 − 9 + 5 = 16
n(C only) = n(C) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C)
= 42 − 8 − 9 + 5 = 30
Total only one = 15 + 16 + 30 = 61
Step 3: At least one = (only one) + (exactly two) + (all three)
= 61 + 12 + 5 = 78
Step 4: None = n(U) − n(A ∪ B ∪ C) = 108 − 78 = 30
Answer: (i) 61 persons, (ii) 12 persons, (iii) 30 persons
Q34). In a study of 139 students, it was found that 72 students study English, 43 study Hindi, 48 study Sanskrit, 8 study English and Hindi, 13 study Hindi and Sanskrit, 8 study English and Sanskrit and 5 study all the three languages. Find the number of students who study: (i) English only (ii) Hindi only (iii) Sanskrit only (iv) English and Hindi only (v) Hindi and Sanskrit only (vi) Sanskrit and English only (vii) none of the three languages
Solution (step-by-step):
n(E∩H)=8, n(H∩S)=13, n(S∩E)=8, n(E∩H∩S)=5.
(A) Pairwise only (subtracting triple overlap):
(E∩H only) = n(E∩H) − n(E∩H∩S) = 8 − 5 = 3
(H∩S only) = n(H∩S) − n(E∩H∩S) = 13 − 5 = 8
(S∩E only) = n(S∩E) − n(E∩H∩S) = 8 − 5 = 3
(B) Only one sets (using formula):
English only = n(E) − n(E∩H) − n(E∩S) + n(E∩H∩S)
= 72 − 8 − 8 + 5 = 61
Hindi only = n(H) − n(E∩H) − n(H∩S) + n(E∩H∩S)
= 43 − 8 − 13 + 5 = 27
Sanskrit only = n(S) − n(S∩E) − n(H∩S) + n(E∩H∩S)
= 48 − 8 − 13 + 5 = 32
(C) Total in union:
Sum = 61 + 27 + 32 + 3 + 8 + 3 + 5 = 139
(D) None = n(U) − n(E ∪ H ∪ S) = 139 − 139 = 0
Final Answers:
(i) English only = 61
(ii) Hindi only = 27
(iii) Sanskrit only = 32
(iv) English & Hindi only = 3
(v) Hindi & Sanskrit only = 8
(vi) Sanskrit & English only = 3
(vii) None = 0
Q35). Two sets A and B have 32 and 18 elements respectively. What is the minimum number of elements in A ∪ B?
Solution:
To minimize n(A ∪ B), maximize overlap (A ⊆ B or B ⊆ A).
⇒ Minimum n(A ∪ B) = max{n(A), n(B)}
= max{32, 18} = 32
Answer: Minimum number of elements in A ∪ B is 32.
Q36). In a group of 280 students, 140 play Hockey (H), 120 play Football (F), and 100 play Volleyball (V). 55 students play both Hockey & Football, 45 play both Football & Volleyball, 30 play both Volleyball & Hockey, and 20 play all three games. Find how many students play at least one of the three games. Also show the counts in each Venn region.
Solution (detailed):
⇒ 55 − 20 = 35
F∩V only = n(F∩V) − n(H∩F∩V)
⇒ 45 − 20 = 25
V∩H only = n(V∩H) − n(H∩F∩V)
⇒ 30 − 20 = 10 Now compute "only" counts (play exactly one sport):
Only H = n(H) − [H∩F only + V∩H only + all three]
⇒ 140 − (35 + 10 + 20) = 75
Only F = n(F) − [H∩F only + F∩V only + all three]
⇒ 120 − (35 + 25 + 20) = 40
Only V = n(V) − [F∩V only + V∩H only + all three]
⇒ 100 − (25 + 10 + 20) = 45 Now sum all seven disjoint regions to get n(H ∪ F ∪ V): Only H (75) + Only F (40) + Only V (45)
+ H∩F only (35) + F∩V only (25) + V∩H only (10) + all three (20)
⇒ 75 + 40 + 45 + 35 + 25 + 10 + 20 = 250 Answer: 250 students play at least one of the three games.
Q37). Out of 60 students in a test: 27 passed Mathematics (M), 32 passed Physics (P), 20 passed Chemistry (C). 12 passed both Mathematics & Physics, 8 passed both Physics & Chemistry, 6 passed both Chemistry & Mathematics, and 4 passed all three subjects. Find the number of students who passed at least one subject, and show the distribution in the Venn diagram.
Solution (detailed):
P∩C only = 8 − 4 = 4
C∩M only = 6 − 4 = 2 Only M = n(M) − [M∩P only + C∩M only + all three]
⇒ 27 − (8 + 2 + 4) = 13
Only P = 32 − (8 + 4 + 4) = 16
Only C = 20 − (4 + 2 + 4) = 10 Sum of all regions = 13 + 16 + 10 + 8 + 4 + 2 + 4 = 57 Answer: 57 students passed at least one subject.
Q38). In a survey it was found that 21 people liked product P1, 26 liked product P2 and 29 liked product P3. If 14 liked products P1 and P2, 12 liked products P2 and P3, 14 liked products P1 and P3, and 8 liked all the three products, find how many liked: (i) product P1 only (ii) product P2 only (iii) product P3 only (iv) exactly two products (v) all the three products (vi) none of the products
Solution:
n(P1∩P2)=14, n(P2∩P3)=12, n(P1∩P3)=14,
n(P1∩P2∩P3)=8.
Pairwise-only:
P1∩P2 only = 14 − 8 = 6
P2∩P3 only = 12 − 8 = 4
P1∩P3 only = 14 − 8 = 6
Singles:
P1 only = 21 − (6+6+8) = 1
P2 only = 26 − (6+4+8) = 8
P3 only = 29 − (6+4+8) = 11
(iv) Exactly two = 6+4+6 = 16
(v) All three = 8
(vi) Union = 1+8+11+6+4+6+8 = 44.
None = (total not given, assume 50) → None = 50 − 44 = 6 (if universe = 50).
Answer: (i)1, (ii)8, (iii)11, (iv)16, (v)8, (vi)6 (assuming 50 people total).
Q39). In a class of 90 students of a school, it was found that 20 study Mathematics, 25 study Physics and 30 study Chemistry, 12 study Mathematics and Physics, 14 study Physics and Chemistry, 10 study Mathematics and Chemistry and 5 study all the three subjects. Find the number of students who study: (i) all the three subjects (ii) Mathematics only (iii) Physics only (iv) Chemistry only (v) exactly two subjects (vi) none of the subjects
Solution:
n(M∩P)=12, n(P∩C)=14, n(M∩C)=10,
n(M∩P∩C)=5, n(U)=90.
Pairwise-only:
M∩P only = 12 − 5 = 7
P∩C only = 14 − 5 = 9
M∩C only = 10 − 5 = 5
Singles:
M only = 20 − (7+5+5) = 3
P only = 25 − (7+9+5) = 4
C only = 30 − (9+5+5) = 11
(i) All three = 5
(ii) M only = 3
(iii) P only = 4
(iv) C only = 11
(v) Exactly two = 7+9+5 = 21
(vi) Union = 3+4+11+7+9+5+5 = 44.
None = 90 − 44 = 46.
Answer: (i)5, (ii)3, (iii)4, (iv)11, (v)21, (vi)46
Q40). In a group of 50 students: 15 study French, 13 Sanskrit, 12 English, 5 French and Sanskrit, 4 English and Sanskrit, 6 English and French and 3 study all the three languages. Find the number of students who study: (i) French only (ii) Sanskrit only (iii) English only (iv) French and Sanskrit only (v) English and Sanskrit only (vi) English and French only (vii) at least one of the three languages
Solution:
n(F∩S)=5, n(E∩S)=4, n(E∩F)=6,
n(F∩S∩E)=3, n(U)=50.
Pairwise-only:
F∩S only = 5 − 3 = 2
E∩S only = 4 − 3 = 1
E∩F only = 6 − 3 = 3
Singles:
F only = 15 − (2+3+3) = 7
S only = 13 − (2+1+3) = 7
E only = 12 − (3+1+3) = 5
(i) F only = 7
(ii) S only = 7
(iii) E only = 5
(iv) F∩S only = 2
(v) E∩S only = 1
(vi) E∩F only = 3
(vii) Union = 7+7+5+2+1+3+3 = 28
Answer: F only=7, S only=7, E only=5, F∩S only=2, E∩S only=1, E∩F only=3, At least one=28
Q41). Out of 100 students, 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 4 in Mathematics and Science, 5 in Science and English and 4 in all the three subjects. Find the number of students who passed in: (i) English and Mathematics but not Science (ii) Mathematics and Science but not English (iii) English and Science but not Mathematics (iv) English only (v) Mathematics only (vi) Science only (vii) in more than one subject only
Solution:
n(E)=15, n(M)=12, n(S)=8,
n(E∩M)=6, n(M∩S)=4, n(S∩E)=5,
n(E∩M∩S)=4.
Pairwise-only:
E∩M only = 6 − 4 = 2
M∩S only = 4 − 4 = 0
S∩E only = 5 − 4 = 1
Singles:
E only = 15 − (2+1+4) = 8
M only = 12 − (2+0+4) = 6
S only = 8 − (0+1+4) = 3
(i) E∩M but not S = 2
(ii) M∩S but not E = 0
(iii) S∩E but not M = 1
(iv) English only = 8
(v) Mathematics only = 6
(vi) Science only = 3
(vii) More than one only = 2+0+1=3
Answer: (i)2, (ii)0, (iii)1, (iv)8, (v)6, (vi)3, (vii)3
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