Let A = {a_1, a_2, a_3, dots, a_n} be a set c | vedclass

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(B) For each element $x \in A$,there are $4$ possibilities regarding its membership in $P$ and $Q$: $x \in P \cap Q$,$x \in P \setminus Q$,$x \in Q \s

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$^nC_2 	imes 3^{n-2}$

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(B) For each element $x \in A$,there are $4$ possibilities regarding its membership in $P$ and $Q$: $x \in P \cap Q$,$x \in P \setminus Q$,$x \in Q \setminus P$,or $x 
otin P \cup Q$.We want $(P - Q)$ to contain exactly $2$ elements.First,we choose $2$ elements out of $n$ to be in $(P - Q)$,which can be done in $^nC_2$ ways.For these $2$ chosen elements,they must be in $P$ but not in $Q$,so there is only $1$ way for each.For the remaining $(n - 2)$ elements,each element can either be in $Q \setminus P$,in $P \cap Q$,or in neither $P$ nor $Q$ (i.e.,$x 
otin P \cup Q$). Thus,there are $3$ choices for each of the remaining $(n - 2)$ elements.Therefore,the total number of ways is $^nC_2 	imes 3^{n-2}$.

Let $A = \{a_1, a_2, a_3, \dots, a_n\}$ 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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