Sets-Relations-And-Functions Question 124
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
Correct Answer: DShow Answer
Answer:
Solution:
$ 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
