A is a set containing n different elements. A subset P of A is chosen. 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
A). $^n{C_3} \times {2^n}$
B). $^n{C_2} \times {3^{n - 2}}$
C). ${3^{n - 2}}$
D). None of these

Hint- This is a problem of combination where we will have to choose the numbers of ways in which the final required set contains only two elements which is common to both subset P and Q of set A. we will use the formula of combination for this, suppose we have to choose any n object from any set having m elements then it is given by $^m{C_{n.}}$.

Complete step- by-step solution -
It is given that the set A has n elements and the number of common elements that set P and Q will have be 2.
First let us remove two elements from the set A and number of ways from which we can do it is
$^n{C_2}$
Let us assume that subset P contains a number of elements
Where $n > a > 0$
And subset Q contains $\left( {n - a - 2} \right)$ number of elements
If we assume subset P contains only one element, therefore the number of ways in which subset P can be selected is $^{n - 1}{C_1}$
The number of possible ways in which the subset Q is selected will be ${2^{n - 3}}$ from the remaining n-3 elements.
Therefore the total possibility if subset P contains 1 element is
=$^{n - 2}{C_1} \times {2^{n - 3}}$
The above we solve for the case if P has only one element, similarly we can solve for the case when the value of is greater than 1 i.e. for 2, 3, 4, …………………n.
Now combining all the cases, we get
$
  { = ^n}{C_2} \times \left( {^{n - 2}{C_0} \times {2^{n - 2}}{ + ^{n - 2}}{C_1} \times {2^{n - 3}} + ............{ + ^{n - 2}}{C_{n - 2}} \times {2^0}} \right) \\
  { = ^{n - 2}}{C_2} \times {\left( {1 + 2} \right)^{n - 2}} \\
  { = ^{n - 2}}{C_2} \times {3^{n - 2}} \\
 $

Hence, the number of ways of choosing P and Q so that $P \cap Q$ contains exactly two elements is $^{n - 2}{C_2} \times {3^{n - 2}}$ and the correct option is B.

Note- In order to solve these types of questions, you need to have a concept of combinations and permutation. In the above question, we subtracted the common elements first and then saw the number of elements that subset P and Q have apart from the common elements and see all the possible combinations.