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Question: A is a set containing n different elements. A subset P of A is chosen. A subset Q of A is again chos...

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∩QP \cap Q contains exactly two elements is
A). nC3×2n^n{C_3} \times {2^n}
B). nC2×3n−2^n{C_2} \times {3^{n - 2}}
C). 3n−2{3^{n - 2}}
D). None of these

Explanation

Solution

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 mCn.^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
nC2^n{C_2}
Let us assume that subset P contains a number of elements
Where n>a>0n > a > 0
And subset Q contains (n−a−2)\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−1C1^{n - 1}{C_1}
The number of possible ways in which the subset Q is selected will be 2n−3{2^{n - 3}} from the remaining n-3 elements.
Therefore the total possibility if subset P contains 1 element is
=n−2C1×2n−3^{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
=nC2×(n−2C0×2n−2+n−2C1×2n−3+............+n−2Cn−2×20) =n−2C2×(1+2)n−2 =n−2C2×3n−2  { = ^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∩QP \cap Q contains exactly two elements is n−2C2×3n−2^{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.