Question

Two finite sets have $m$ & $n$ elements, if the total number of subsets of the first set is 56 more than the total number of subsets of the second. The value of $m$ & $n$ are:
$
  A.{\text{ }}7,6 \\
  B.{\text{ }}6,3 \\
  C.{\text{ }}5,1 \\
  D.{\text{ }}8,7 \\
 $

Answer 1

Hint- For any set with a given $x$ number of elements, the total number of subsets that can be formed from that set is ${2^x}$ . Use this property to reach the answer.

Let set $A$ has $m$ number of elements.
And let set $B$ has an $n$ number of elements.
As we know that for any set with $x$ number of elements, the total number of subsets is ${2^x}$.
Total number of subsets of $A = {2^m}$ .
Total number of subsets of $B = {2^n}$ .
According to the question number of subsets of A is 56 more than that of B
$ \Rightarrow {2^m} - {2^n} = 56$
Taking ${2^n}$ common from the LHS
$ \Rightarrow {2^n}\left( {{2^{m - n}} - 1} \right) = 56$
So from the above equation we have
${2^n}$ is even.[power of 2]
${2^{m - n}} - 1$ is odd. [power of 2 subtracted by one]
56 can be simplified as a product of odd and even as $56 = 8 \times 7$
Now,
$
  56 = 8 \times 7 = {2^3} \times 7 \\
   \Rightarrow {2^n}\left( {{2^{m - n}} - 1} \right) = {2^3} \times 7 \\
   \Rightarrow n = 3 \\
$
Now solving for $m$ with the help of the second term.
$
  8\left( {{2^{m - 3}} - 1} \right) = 8 \times 7 \\
   \Rightarrow {2^{m - 3}} - 1 = 7 \\
   \Rightarrow {2^{m - 3}} = 7 + 1 = 8 \\
   \Rightarrow {2^{m - 3}} = {2^3} \\
 $
Now comparing the powers of both the side, we get
$
   \Rightarrow m - 3 = 3 \\
   \Rightarrow m = 6 \\
$
So, we have $m = 6$ & $n = 3$ .
Hence, option B is the correct option.

Note- For solving questions related to sets, basic properties like number of subsets to be formed is very important and must be remembered. The above equation had 2 unknown variables and only one equation. In order to solve such equations try to use the practical aspect of the question as in the above case we have considered that the number of subsets will be integer.