Question

$A$ is a set containing $n$ 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. Let ${}^n{C}_2\times k^{n-2}$ be the number of ways of choosing $P$ and $Q$ so that $P\cap Q$ contains exactly two elements.Find $k$ ?

Answer 1

Let $A=\{a_1,a_2,a_3,...,a_n\}$
The two elements $P$ and $Q$ such that $P\cap Q$ can be chosen out of $n$ is ${}^n{C}_2$ ways a general element of $A$ must satisfy one of the following possibilities : (here general element be $a_i(1\le i\le n)$)
$\left(i\right)a_i\in P$ and $a_i\in Q$

$\left(ii\right)a_i\in P$ and $a_i\notin Q$
$\left(iii\right)a_i\notin P$ and $a_i\in Q$

$\left(iv\right)a_i\notin P$ and $a_i\notin Q$
Let $a_1,a_2\in P\cap Q$
there is only one choice each of them (i.e. (i) choice) and three choices (ii), (iii) and (iv) for each of remaining $(n-2)$ elements.
$\therefore$ No. of ways of remaining elements $=3^{n-2}$
Hence, the no. of ways in which $P\cap Q$ contains exactly two elements $={}^n{C}_2\times 3^{n-2}$