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(B) For each element $x \in A$,there are $4$ possibilities regarding its membership in subsets $P$ and $Q$: $1$. $x \in P$ and $x \in Q$ $2$. $x \in P

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$^nC_2 \cdot 3^{n-2}$

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(B) For each element $x \in A$,there are $4$ possibilities regarding its membership in subsets $P$ and $Q$: $1$. $x \in P$ and $x \in Q$ $2$. $x \in P$ and $x 
otin Q$ $3$. $x 
otin P$ and $x \in Q$ $4$. $x 
otin P$ and $x 
otin Q$ The condition $(P - Q)$ contains exactly $2$ elements means that for exactly $2$ elements of $A$,the condition $x \in P$ and $x 
otin Q$ must hold. There are $^nC_2$ ways to choose these $2$ elements. For the remaining $(n - 2)$ elements,each element can be in any of the other $3$ states (i.e.,$x \in P \cap Q$,$x \in Q \setminus P$,or $x 
otin P \cup Q$). Thus,the total number of ways is $^nC_2 \cdot 3^{n-2}$.

Let $A = \{a_1, a_2, a_3, ..., a_n\}$ be a set containing $n$ elements. Two subsets $P$ and $Q$ of $A$ are formed independently. The number of ways in which these subsets can be formed such that $(P - Q)$ contains exactly $2$ elements is:

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