Consider the set A1, A2, A3 of | A1 | = 15, | A2 | = 12, | A3 | = 6, $| A1 \cap A2 \cap A3 | =3$ What is the minimum and maximum value of $|A1\cup A2 \cup A3|$?
My attempt: the minimum value is 15 Because A1 is biggest set can contain 9 from A2 + 3 from A3 and 3 from $| A1 \cap A2 \cap A3 | =3$ so all satisfy
But for biggest possible value
$| A1 | + |A2 | + |A3 | - (|A1 \cup A2 | + |A2 \cup A3| + |A3 \cup A1) +3 $
Suppose a is $|A1 \cup A2 |$ and b= $|A2 \cup A3| $ and c= $|A3 \cup A1|$
$15+12+6-(a+b+c)+3=$
$a +b+c =36$
But we know also that
$a+b \leq 12$
$a+c \leq 9$
$b+c \leq 3$
Sum above $ 2a+2b+2c=24$
So $a+b+c=12$
Then how can i conclude what is the biggest value that is possible?
Consider the sets as people and there are 3 circle for every sets in venn diagram
What you have written is incorrect. You say, for example, that we know $a+b\leq12$ where $a=\mid A_1\cup A_2\mid$ and $b=\mid A_2\cup A_3\mid.$ On the contrary, since $\mid A_2\mid =12,$ we know $a+b\geq12$.
The only duplication required is that there are $3$ elements that belong to all $3$ sets. Then just make all other elements distinct. For example, $A_1$ should have $12$ elements that belong to neither $A_2$ nor $A_3$. You should find $27$ elements is the maximum.