Cardinality of both following sets are $3^n$, but I can not understand how:
$$ {\{(S,T) \mid S \in \mathcal{P}(A),\, T \in \mathcal{P}(A),\, S \cap T = \emptyset\}}$$ $$ {\{(S,T) \mid S \in \mathcal{P}(A),\, T \in \mathcal{P}(A),\, S \cup T = A\}}$$
$\mathcal{P}$ indicates power set and $A = \{1,2,3, \dots ,n\}$. I really appreciate if anyone can answer my question. Thanks a lot.
For the first set, \begin{align} {\{(S,T) | S \in P(A), T \in P(A), S \cap T = \oslash\}} &= {\{(S,T) | S \in P(A), T \subset S^c\}} \\ &= \bigcup_{S \subset A} \bigcup_{T \subset S^c} {\{ (S,T) \}} \end{align} where the two unions are disjoint so the cardinal of this set is $$ \sum_{S \subset A}\sum_{T \subset S^c}1= \sum_{S \subset A} 2^{n-card(A)}=2^n\sum_{k=0}^n\binom{n}{k} \left( \frac{1}{2} \right)^k = 3^n.$$ For the second set it is the same idea : \begin{align} {\{(S,T) | S \in P(A), T \in P(A), S \cup T = A\}} &= {\{(S,T) | S \in P(A), A\setminus S \subset T\}} \\ &= \bigcup_{S \subset A} ~~ \bigcup_{T:A \setminus S \subset T}~~ {\{(S,T)\}} \end{align} and for the cardinal, $$ \sum_{S \subset A} ~~\sum_{T: A\setminus S \subset T}~~1= \sum_{S \subset A} 2^{card(S)} = \sum_{k=0}^n \binom{n}{k}2^k=3^n.$$