Question

Two finite sets have $m$ and $n$ elements. The total number of subsets of the first set is $16$ more than the total number of subsets of the second set. Then ${{m}^{2}}-{{n}^{2}}=$
(1). $1$
(2). $9$
(3). $20$
(4). $5$

Answer 1

Here we will assume the number of subsets for the two finites sets having $m$ and $n$ elements as $x$ and $y$. Now we will calculate the number of subsets for the finite sets $m$ and $n$ elements by using the formula ${{2}^{a}}$, where $a$ is the number of elements in the set. Then we will have the values of $x$ and $y$. In the problem they have mentioned that “The total number of subsets of the first set is $16$ more than the total number of subsets of the second set”, from this statement we can establish the relation between $x$ and $y$. Now we are going to calculate the values of $m$ and $n$ by using the above relation. After getting the values of $m$ and $n$, we can simply find the value of ${{m}^{2}}-{{n}^{2}}$.

Complete step-by-step answer:
Given that, Two finite sets have $m$ and $n$ elements.
Let the number of subsets for both finite sets are $x$ and $y$.
We know that the number of subsets for a finite set having $a$ number of elements is given by ${{2}^{a}}$.
Hence the values of $x$ and $y$ are
$x={{2}^{m}}$ and $y={{2}^{n}}$
In the problem they have mentioned that the total number of subsets of the first set is $16$ more than the total number of subsets of second set, then
$\begin{aligned} & x=y+16 \\\ & x-y=16 \\\ \end{aligned}$
Substituting the value of $x$ and $y$, then we will get
$\begin{aligned} & x-y=16 \\\ & {{2}^{m}}-{{2}^{n}}=16 \\\ \end{aligned}$
Taking ${{2}^{n}}$ as common from the term ${{2}^{m}}-{{2}^{n}}$, then we will have
${{2}^{n}}\left( {{2}^{m-n}}-1 \right)=16...\left( \text{i} \right)$
Here we can clearly say that ${{2}^{n}}$ is Even number and ${{2}^{m-n}}-1$ is Odd number. So we need to factorize the number $16$ as the product of one Even number and one Odd number to get the values of $m$ and $n$.
So,
$\begin{aligned} & 16=2\times 8 \\\ & 16=2\times 2\times 4 \\\ & 16=2\times 2\times 2\times 2 \\\ & 16={{2}^{4}}\times 1 \\\ \end{aligned}$
Substituting the value of $16$ in equation $\left( \text{i} \right)$, then we will get
${{2}^{n}}\left( {{2}^{m-n}}-1 \right)={{2}^{4}}\times 1$
Equating on both sides, we will have
$n=4$ ,
$\begin{aligned} & {{2}^{m-n}}-1=1 \\\ & \Rightarrow {{2}^{m-n}}=2 \\\ & \Rightarrow {{2}^{m-4}}={{2}^{1}} \\\ & \Rightarrow m-4=1 \\\ & \Rightarrow m=5 \\\ \end{aligned}$
Hence the values of $m$ and $n$ are $5,4$ respectively.
Now the value of ${{m}^{2}}-{{n}^{2}}$ is
$\begin{aligned} & {{m}^{2}}-{{n}^{2}}={{5}^{2}}-{{4}^{2}} \\\ & =25-16 \\\ & =9 \\\ \end{aligned}$
Hence the value of ${{m}^{2}}-{{n}^{2}}=9$.
Second Option is the correct One.

So, the correct answer is “Option (2)”.

Note: This problem requires spontaneous reactions of students at every point. Students may stop their process at factorizing the value $16$ because there is no Odd number in factors of $16$, but you should remember that $1$ is the factor of every number and it can be treated as an Odd number. Then only you can move further.