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 $2^{m+n-2k}-2^{m-k}-2^{n-k}$

this doesn't work, you have more options then you mentioned

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 Originally Posted by eden 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. Originally Posted by SlipEternal 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? Originally Posted by Plato 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 $\left|\mathcal{P}(A\cup B)-[\mathcal{P}(A)\cup\mathcal{P}(B)]\right|=2$
Sorry, $|p(A\cup B)| = 2^{m+n-k}$. So, $|p(A\cup B) - (p(A)\cup P(B))| = 2^{m+n-k}-2^m-2^n+2^k$
In your case, we have $2^{2+2-1}-2^2-2^2+2^1 = 2$, just as you found. I just miswrote the formula. You understood the question.