0
$\begingroup$

I tried using a Venn diagram to understand this property but when I do so the middle portion $(A \cap B \cap C)$ repeats four times. So can Venn diagrams not be used for $n()$ properties? So is the above rule not the same as

$$(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)?$$

And if both are same and valid, can someone explain the Venn diagram?

$\endgroup$
3
$\begingroup$

It does work... Perhaps you over looked something?

enter image description here

To get the shaded set with no overlaps, you need to add the three sets $A$, $B$ and $C$. That will give you overlaps in $A \cap B$, $B \cap C$ and $C \cap A$, and 2 overlaps in $A \cap B \cap C$. To get rid of these overlaps we subtract $A \cap B$, $B \cap C$ and $C \cap A$, but now that means we are missing $A \cap B \cap C$. Now we add $A \cap B \cap C$ to get the result

$$n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(C \cap A)+n(A \cap B \cap C)$$

$\endgroup$
2
$\begingroup$

Start with the simpler case of $A \cup B$.

If you verify that $|A \cup B|=|A|+|B|-|A \cap B|$, you can apply it to $(A \cup B) \cup C$.

And see Inclusion–exclusion principle.


With two sets, the intuition is to remove the elements counted twice, i.e. the elements that belong to both $A$ and $B$.

With three sets, the case is similar : we start removing $A \cap C$, $B \cap C$ and $A \cap C$.

But in this way we have removed thrice the elements that belongs to all three sets, and this is wrong because we have removed all their occurrences.

Tis is the reason why we have to add at the counting the elements belonging to $A \cap B \cap C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.