... The set P is reconstructed by replacing its elements. A subset 'B' of P is again selected at random. What is the number of ways of selecting the subsets such that $A∩B=\phi$ And what is the number of unordered pairs $(A, B)$ such that $A∩B=\phi$?
So I figured out that if we select r elements out of the n and assign them to A, we will have $n_{C_r}$ ways. To select elements for B such that it has no elements in common with A, we will select the subsets of P with $n-r$ elements. So that gives us total number of ways $= \sum_{0}^{n}{n \choose r}\cdot(2^{n-r})$ which is the expansion for $(2+1)^n =3^n$.
I'm stuck at finding the number of unordered pairs. Can someone please help me with the approach I'm supposed to use for this?
The number of ways of selecting the subsets $A$ and $B$ such that $A \cap B = \varnothing$ is the number of ways of dividing the elements of $P$ into three buckets, which we can label $A$, $B$, and $N$ (for Neither). Each element of $P$ ends up in one of $3$ possible buckets and each different division of elements into buckets yields a different ordered pair, so the number of choices for ordered pairs of $A$ and $B$ is $3^n$. That saves you the trouble of computing the sum. Your analysis, though, is also correct.
Since every ordered pair $(A, B)$ except for $(\varnothing, \varnothing)$ can be matched with the corresponding ordered pair $(B, A)$, there are $1+ \frac{3^n-1}{2}$ unordered pairs.