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
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
Updated On: Mar 10, 2025
- 26
- 27
- 28
- Any of the three values 26, 27, 28 is possible
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The Correct Option is D
Solution and Explanation
$n(A)=10, n(B)=15, n(20), n(A \cap B)=8, n(B \cap C)=9$
possible value of $n$ (AUBUC)
$ n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B) -n(B \cap C)-n(A \cap C) +n(A \cap B \cap C) $
$=10+15+20-8-9-n(A \cap C)+n(A \cap B \cap C)$
$=28-n(A \cap C)+n(A \cap B \cap C)$
We see here,
$28-n(A \cap C)+n(A \cap B \cap C) \geqslant 0$
$\Rightarrow n(A \cup B \cup C) \leqslant 28\,\,\,...(1)$
We see, $n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=10+15-8$
$=17$
$n(B \cup C) =n(B)+n(C)-n(B \cap C)$
$ =15+20-9 $
$=26 $
Obviously, $n(A \cup B \cup C) \geqslant 26$ also,
$n(A \cup B \cup C) \geqslant 17\,\,\, ...(2)$
$\therefore$ from (1) & (2)
$26 \leqslant n(A \cup B \cup C) \leqslant 28$
possible value of $n$ (AUBUC)
$ n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B) -n(B \cap C)-n(A \cap C) +n(A \cap B \cap C) $
$=10+15+20-8-9-n(A \cap C)+n(A \cap B \cap C)$
$=28-n(A \cap C)+n(A \cap B \cap C)$
We see here,
$28-n(A \cap C)+n(A \cap B \cap C) \geqslant 0$
$\Rightarrow n(A \cup B \cup C) \leqslant 28\,\,\,...(1)$
We see, $n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=10+15-8$
$=17$
$n(B \cup C) =n(B)+n(C)-n(B \cap C)$
$ =15+20-9 $
$=26 $
Obviously, $n(A \cup B \cup C) \geqslant 26$ also,
$n(A \cup B \cup C) \geqslant 17\,\,\, ...(2)$
$\therefore$ from (1) & (2)
$26 \leqslant n(A \cup B \cup C) \leqslant 28$
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