Question

$A$ is a set containing $n$ elements. A subset $P_1$ of $A$ is chosen. The set $A$ is reconstructed by replacing all the elements of $P_1$. Next, a subset $P_2$ of $A$ is chosen and again reconstructed by replacing all the elements of $P_2$. In this way, $m>1$ subsets $P_1,P_2,P_3,\cdots ,P_m$ of $A$ are chosen. The number of ways of choosing $P_1,P_2,P_3\cdots ,P_m$ is

• A. $(2^m-1{)}^n\text{if}{P}_1\cap P_2\cap \cdots \cap P_m=\varphi$
• B. $\left(2^{mn}\right)\text{if}{P}_1\cup P_2\cup \cdots \cup P_m=A$
• C. $\left(2^{mn}\right)\text{if}{P}_1\cap P_2\cap \cdots \cap P_m=\varphi$
• D. $(2^m-1{)}^n\text{if}{P}_1\cup P_2\cup \cdots \cup P_m=A$

Answer 1

Answer: (A), (D)

The correct options are
A $(2^m-1{)}^n\text{if}{P}_1\cap P_2\cap \cdots \cap P_m=\varphi$
D $(2^m-1{)}^n\text{if}{P}_1\cup P_2\cup \cdots \cup P_m=A$

Let $A=\{a,a_1,...,a_n\}$
$\forall a_i(1\le i\le n),a_i\in P_j\text{or}{a}_i\notin P_j(1\le j\le m)$
i.e., there are $2^m$ choices in which $a_i(1\le i\le n)$ may belong to $P_j$'s.
Out of these, there is only choice in which $a_i\in P_j\forall j=1,2,...,m$
which is not favourable for
$P_1\cap P_2\cap \cdots \cap P_m=\varphi$
Thus $a_i\notin P_1\cap P_2\cap \cdots \cap P_m$ in $(2^m-1)$ ways.
$\because$ There are total $n$ elements in set $A$, the total number of choices is $(2^m-1{)}^n$

Also, there is exactly one choice, in which $a_i\notin P_j\forall j=1,2,...,m$
which is not favourable for $P_1\cup P_2\cup \cdots \cup P_m=A$
Thus, $a_i$ can belong to $P_1\cup P_2\cup \cdots \cup P_m$ in $2^m-1$ ways.
$\because$ There are $n$ elements in set $A$, the number of ways in which $P_1\cup P_2\cup \cdots \cup P_m$ can be equal to $A$ is $(2^m-1{)}^n$