Given sets A and B, so that |A|=m and |B|=n and |A∩B|=k, and p(S) the Power Set of S, what is the value of |p(A∪B)-(p(A) ∪ p(B)) |?
how do i solve it.
What is in $p (A\cup B)$ that is not in $p (A)\cup p (B) $
That's gonna be $p(A\Delta B)-p (A-B)-p (B-A) $ where $\Delta $ is symmetric difference.
$|A\Delta B|=|A-B|+|B-A|=(m-k)+(n-k) $
Can you finish from here?
I've not checked my work, but I think the final answer is $2^{m+n-2k}-2^{m-k}-2^{n-k} $
You are right. Let's look at sets in $p (A\cup B) $ and ones in $p (A)\cup p (B) $
So, consider the intersection of $p (A)\cap p (B) $. It is precisely $p (A\cap B) $
So $|p (A\cup B)-(p (A)\cup p (B))|-|p (A\cup B)|-|p (A)|-|p (B)|+|p (A\cap B)|=2^{m+n}-2^m-2^n+2^k $ by Inclusion/Exclusion
Let $A=\{a,b\}~\&~B=\{b,c\}$ thus $m=n=2~\&~k=1$
$\mathcal{P}(A\cup B)-[\mathcal{P}(A)\cup\mathcal{P}(B)]=\{\{a,c\},\{a,b,c\}\}$
In this example $\left|\mathcal{P}(A\cup B)-[\mathcal{P}(A)\cup\mathcal{P}(B)]\right|=2$
Did I not understand the question?