Question

$A$ is a set containing $n$ elements. A subset $P_1$ 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 $P_1,P_2,....,P_m$ are choosen. The number of ways of choosing $P_1,P_2,\cdots ,P_m$ 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)

$(2^m-1{)}^n$ if $P_1\cup P_2\cup ...\cup P_m=A$

Let $A=a_1,a_2,....,a_n.$ For each $a_i(1\le i\le n),$ we have either $a_i\in P_j\text{or}{a}_i\notin P_j).$ That is, there are $2$ choices in which $a_i(1\le i\le n)$ may belong to the $P_j$'s. One of these, there is only one choice, in which $a_i\in P_j$ for all $j=1,2,...,m$ which is not favourable for $P_1\cap P_2\cap ....\cap P_m$ to be $\varphi .$ Thus, $a_i,\notin P_1\cap P_2\cap ....\cap P_m$ 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, $a_1\notin P_j$ for all $j=1,2,...,m$ which is not favourable for $P_1\cup P_2\cup ....\cup P_m$ to be equal to $A$.
Thus, $a_i$ can belong to $P_1\cup P_2\cup ...\cup P_m$ in $(2^m-1)$ ways.
Since there are $n$ elements in set $A$, the total number of choices $(2^m-1{)}^n.$