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+1 vote

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. $22$
in Set Theory & Algebra by Boss (17.5k points)
recategorized by | 49 views
$8$ should be the answer.

2 Answers

+2 votes
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.
by Boss (19.1k points)
edited by

@jeet nice analysis.

But i have a doubt did you arrive at the sets A,B,C



I have just taken a simplified test case to make it easy.

You can't have an example set more simplified than this.

So, it's just an example that can be as simplified as possible.

There is another way to analyze this but I found this as the easiest.

Thanks for the reply... Can you explain other method... Because i feel it would be difficult to get to a example in exam.

Okay, So observe the below formula carefully.

$\mathrm {A \cup B \cup C = A + B + C - A\cap B-B\cap C -C \cap A + A\cap B\cap C}$

For, $\mathrm {A \cup B\cup C}$ is minimum when the negative terms in the above formula are maximum.

You can analyze this way.

Check out the below two links and let me know.

+1 vote

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}. 


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 $|A\cap B\cap C|=7$.

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



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 $A\cap C$

by Active (4.4k points)

@jeet  can you verify this answer for mistakes..

This is correct and a good explanation as well.
Thank you :)
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