Question

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

• A. $(2^m-1{)}^n$ if $P_1\cap P_2\cap ...\cap P_m=\varphi$
• B. $2^{mn}$ if $P_1\cup P_2\cup ...\cup P_m=A$
• C. $2^{mn}$ if $P_1\cap P_2\cap ...\cap P_m=\varphi$
• D. $(2^m-1{)}^n$ if $P_1\cup P_2\cup ...\cup P_m=A$

Answer 1

Answer: (A), (D)

The correct options are
A $(2^m-1{)}^n$ if $P_1\cap P_2\cap ...\cap P_m=\varphi$
D $(2^m-1{)}^n$ if $P_1\cup P_2\cup ...\cup P_m=A$

Let $A=a_1,a_2...a_n$
Now $a_i$ may belong to $P_j$ or $a_i$ may not belong to $P_j$.
That is, there are $2^m$ choices in which $a_i$ may belong to $P_j$.
Out of these, there is only one choice in which $a_iϵP_j$.
Thus $a_i\mathrm{ϵ/}{P}_1\cap P_2...P_m$ in $2^m-1$ ways.
Since there are n elements in set $A$, the total number of choices is $(2^m-1{)}^n$
Also, there is exactly one choice, in which $a_i\mathrm{ϵ/}{P}_j$ which is not favorable for $P_1\cup P_2\cup ...P_m$ to be equal to $A$.
Thus, $a_iϵP_1\cup P_2...P_m$ in $2^m-1$ ways.
Since, there are $n$ elements in set $A$, the number of ways in which $P_1\cup P_2..P_m$ will be $(2^m-1{)}^n$.