Question

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

Answer 1

The correct option is B ${}^n{C}_2\times 3^{n-2}$

The two common elements can be selected in ${}^n{C}_2$ ways. Remaining $n-2$ elements, each can be chosen in three ways, i.e., $a\in P$ and $a\text{/}\in Q$ or $a\in P$ or a is neither in P nor in Q.Therefore, the total number of ways is ${}^n{C}_2\times 3^{n-2}$.