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Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) is
- a)26
- b)27
- c)28
- d)Any of the three values 26, 27, 28 is possible
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C)...
We have
n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n(B∩C) – n (C ∩ A) + n (A∩B ∩ C)
= 10 +15 + 20 – 8 – 9 – n (C ∩ A) + n (A ∩ B ∩C)
= 28 – {n(C ∩ A) – n (A ∩ B ∩ C)} ...(i)
Since n (C ∩ A) ≥ n (A ∩ B ∩ C)
We have
n (C ∩ A) – n (A ∩ B ∩ C) ≥ 0 ...(ii)
From (i) and (ii)
n (A ∪ B ∪ C) ≤ 28 ...(iii)
Now, n(A ∪ B) = n (A) +n (B) – n (A Ç B) = 10 + 15 – 8 = 17
and n (B ∪ C) = n (B) + n (C) – n (B Ç C) = 15 + 20 – 9 = 26
Since, n (A ∪ B ∪ C) ≥ n (A∪C) and n (A∪B∪C) ≥ n (B∪C),
we have n (A∪B∪C) ≥ 17 and n (A∪B∪C) ≥ 26
n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n(B∩C) – n (C ∩ A) + n (A∩B ∩ C)
= 10 +15 + 20 – 8 – 9 – n (C ∩ A) + n (A ∩ B ∩C)
= 28 – {n(C ∩ A) – n (A ∩ B ∩ C)} ...(i)
Since n (C ∩ A) ≥ n (A ∩ B ∩ C)
We have
n (C ∩ A) – n (A ∩ B ∩ C) ≥ 0 ...(ii)
From (i) and (ii)
n (A ∪ B ∪ C) ≤ 28 ...(iii)
Now, n(A ∪ B) = n (A) +n (B) – n (A Ç B) = 10 + 15 – 8 = 17
and n (B ∪ C) = n (B) + n (C) – n (B Ç C) = 15 + 20 – 9 = 26
Since, n (A ∪ B ∪ C) ≥ n (A∪C) and n (A∪B∪C) ≥ n (B∪C),
we have n (A∪B∪C) ≥ 17 and n (A∪B∪C) ≥ 26
Hence n (A∪B∪C) ≥ 26 ...(iv)
From (iii) and (iv)
we obtain 26 ≤ n (A∪B∪C)≤ 28
Also n (A∪B∪C) is a positive integer
∴ n(A∪B∪C) = 26 or 27 or 28
From (iii) and (iv)
we obtain 26 ≤ n (A∪B∪C)≤ 28
Also n (A∪B∪C) is a positive integer
∴ n(A∪B∪C) = 26 or 27 or 28
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Community Answer
Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C)...
To find the maximum possible value for n(A ∩ B ∩ C), we need to find the smallest possible size of the intersection of sets A, B, and C.
By the inclusion-exclusion principle, we have:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
Given that n(A) = 10, n(B) = 15, n(C) = 20, and n(A ∪ B ∪ C) = 30, we can substitute these values into the equation:
30 = 10 + 15 + 20 - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
Simplifying this equation, we get:
30 = 45 - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
Rearranging the terms, we have:
n(A ∩ B ∩ C) = n(A ∩ B) + n(A ∩ C) + n(B ∩ C) - 15
To maximize n(A ∩ B ∩ C), we need to minimize the values of n(A ∩ B), n(A ∩ C), and n(B ∩ C). Since A, B, and C are finite sets, the minimum possible size for each intersection is 0. Therefore, the maximum possible value for n(A ∩ B ∩ C) is:
n(A ∩ B ∩ C) = 0 + 0 + 0 - 15 = -15
However, the size of a set cannot be negative. Therefore, the maximum possible value for n(A ∩ B ∩ C) is 0.
By the inclusion-exclusion principle, we have:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
Given that n(A) = 10, n(B) = 15, n(C) = 20, and n(A ∪ B ∪ C) = 30, we can substitute these values into the equation:
30 = 10 + 15 + 20 - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
Simplifying this equation, we get:
30 = 45 - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
Rearranging the terms, we have:
n(A ∩ B ∩ C) = n(A ∩ B) + n(A ∩ C) + n(B ∩ C) - 15
To maximize n(A ∩ B ∩ C), we need to minimize the values of n(A ∩ B), n(A ∩ C), and n(B ∩ C). Since A, B, and C are finite sets, the minimum possible size for each intersection is 0. Therefore, the maximum possible value for n(A ∩ B ∩ C) is:
n(A ∩ B ∩ C) = 0 + 0 + 0 - 15 = -15
However, the size of a set cannot be negative. Therefore, the maximum possible value for n(A ∩ B ∩ C) is 0.
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Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer?
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Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer?.
Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer?.
Solutions for Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let A, B, C be finite sets. Suppose that n (A) = 10, n (B) = 15, n (C) = 20, n(A ∩ B) = 8 and n(B ∩ C) = 9. Then the possible value of n (A ∪ B ∪ C) isa)26b)27c)28d)Any of the three values 26, 27, 28 is possibleCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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