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'A' is a set containing 'n' different elements. A subset P of 'A' is chosen. The set 'A' is reconstructed by replacing the elements of P. A subset 'Q' of 'A' is again chosen. The number of ways of choosing P and Q so that $P \cap Q$ contains exactly two elements is

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$^{n} C_{3} \cdot 2^{n}$

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

$3^{n - 2}$

none of these

answer is B.

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Detailed Solution

Two elements for set $P \cap Q$ can be selected in $^{n} C_{2}$ ways. Each of the remaining (n - 2) elements can be put in any of the three sets $(P \cap Q^{'}), (P^{'} \cap Q) or (P \cup Q)^{'}$ .

So, total number of subsets $^{n} C_{2} \times 3^{n - 2}$

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