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You are given three sets $A,B,C$ in such a way that 

  1. the set $B \cap C$ consists of $8$ elements,
  2. the set $A\cap B$ consists of $7$ elements, and
  3. the set $C\cap A$ consists of $7$ elements.

The minimum number of elements in the set $A\cup B\cup C$ is

  1. $8$
  2. $14$
  3. $15$
  4. $22$
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2 Answers

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Answer: $\textbf A$

Let $$\mathrm A = \{1, 2, 3, 4, 5, 6, 7\}$$

$$\mathrm B=\{1, 2, 3, 4, 5, 6, 7, 8\}$$

$$\mathrm C=\{1, 2, 3, 4, 5, 6, 7, 8\}$$

Now, it is given that: $$\mathrm B\cap \mathrm C = 8\;\text{elements}$$

$$\mathrm A\cap \mathrm B = 7\;\text{elements}$$

$$\mathrm C \cap \mathrm A = 7\;\text{elements}$$

 

$\therefore $ Minimum number of elements in $\mathrm A\cup \mathrm B\cup \mathrm C = 8\;\text {elements}$

Hence, option $\textbf A$ is the right answer.
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We know that,$A\cup B\cup C=A+B+C-A\cap B-A\cap C-B\cap C+A\cap B\cap C$.

For $|A\cup B\cup C|$ to be minimum,$|A|,|B|,|C|,$$|A\cap B\cap C|$ has to be minimum and $|A\cap B|,|B\cap C|, |A\cap C|$ has to be maximum.

So let us assume that $B\cap C$={1,2,3,4,5,6,7,8}.So the elements 1,2,3,4,5,6,7,8 has to be in both $B$,$C$  for sure.

Now B can be of the form $B$={1,2,3,4,5,6,7,8....} which can be visualized as B={1,2,3,4,5,6,7,8} $\cup$ $X$.

where X is the set of remaining elements of $B$ apart from {1,2,3,4,5,6,7,8}.Here X can be finite or infinite.

Now since $|B|$ has to be  minimum, $|X|$ has to be zero i.e.,$X$ has to be a  empty set.

$B=${1,2,3,4,5,6,7,8} $\cup$ $X$$=${1,2,3,4,5,6,7,8}. 

i.e.,$B$={1,2,3,4,5,6,7,8}

Thinking in same manner,$C$={1,2,3,4,5,6,7,8}.

Now coming to $|A\cap C|$$=7$.

since $C$={1,2,3,4,5,6,7,8}, we can assume $A\cap C$={1,2,3,4,5,6,7}

Hence we can assume $A$={1,2,3,4,5,6,7....}={1,2,3,4,5,6,7} $\cup$ $X$

where $X$ is the set of remaining elements of $A$ apart from {1,2,3,4,5,6,7,8}.

Now since $|A|$ has to be  minimum, $|X|$ has to be zero i.e.,$X$ has to be a  empty set.

$\therefore$ $A$={1,2,3,4,5,6,7}

Hence $A\cap B\cap C=${1,2,3,4,5,6,7}. and $|A\cap B\cap C|=7$.

$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| +|A\cap B\cap C|$.

=7+8+8-7-7-8+7

=8


Here, $A\cap C$ can be any of {1,2,3,4,5,6,7} or {1,2,3,4,5,6,8} or {1,2,3,4,5,7,8} or {1,2,3,4,6,7,8} or {1,2,3,5,6,7,8} or{1,2,4,5,6,7,8} or {1,3,4,5,6,7,8} or {2,3,4,5,6,7,8}.

Just remove one element from $C$.That will give you $A\cap C$

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