Sets Relations And Functions Question 109
Question: Let A, B, C be finite sets. Suppose that $ n(A)=10, $ $ n(B)=15, $ $ n(C)=20, $ $ n(A\cap B)=8 $ and $ n(B\cap C)=9. $ Then the possible value of $ n(A\cup B\cup C) $ is
Options:
A) 26
B) 27
C) 28
D) Any of the three values 26, 27, 28 is possible
Show Answer
Answer:
Correct Answer: D
Solution:
- [d]We have
$ n(A\cup B\cup C)=n(A)+n(B)+n(C)- $
$ n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C) $
$ =10+15+20-8-9-n(C\cap A)+n(A\cap B\cap C) $
$ =28-{n(C\cap A)-n(A\cap B\cap C)} $ (i) Since $ n(C\cap A)\ge n(A\cap B\cap C) $
We have $ n(C\cap A)\ge n(A\cap B\cap C)\ge 0 $ (ii) From (i) and (ii): $ n(A\cup B\cup C)\le 28 $ (iii) Now, $ n(A\cup B)=n(A)+n(B)-n(A\cap B)=10+15-8=17 $ and $ n(B\cup C)=n(B)+n(C)-n(B\cap C)=15+20-9=26 $ since, $ n(A\cup B\cup C)\ge n(A\cup C) $ and $ n(A\cup B\cup C)\ge n(B\cup C) $ we have $ n(A\cup B\cup C)\ge 17 $ and $ n(A\cup B\cup C)\ge 26 $
Hence $ n(A\cup B\cup C)\ge 26 $ (iv) From (iii) and (iv) we obtain $ 26\le n(A\cup B\cup C)\le 28 $
Also $ n(A\cup B\cup C) $ is a positive integer
$ \therefore ,n(A\cup B\cup C)=26or27or28 $
BETA
