Question

If A and B are two finite sets such that n(A) >n(B) and the difference of the number of
elements of the power sets of A and B is 96, then n(A) -n(B) =................….…​

Answer 1

$\red{\frak{Given}} \begin{cases} \textsf{ A and B are two finite sets . }\\\textsf{ n(A) > n(B) }.\end{cases}$

$\red{\frak{To \ Find}} \begin{cases} \textsf{ The value of \textbf{ n(A) - n(B) } .}\\ \textsf{ The value of \textbf{ n(A) + n(B) } .}\end{cases}$

According to Question , A and B are finite sets and the number of elements in set A is Greater than the number of elements in set B than 96.

We know that if we have a Set X such that the number of elements in set X is k , then the total number of Power sets is given by , $\sf 2^k$ .

• L E T :-

$\qquad\qquad\sf \odot\:\: n_{(Elements \ in \ Set \ A )}= \textsf{\textbf{m}}. \\\qquad\qquad\sf \odot\:\; n_{(Elements \ in \ Set \ B )}= \textsf{\textbf{n}}.$

$\red{\bigstar}\underline{\textsf{ According to Question :- }}$

$\sf:\implies\pink{ n(A)-n(B) = 96 }.\\\\\sf:\implies 2^{m}- 2^n = 32 \times 3 \\\\\sf:\implies 2^m -2^n = (2^5)(3) \\\\\sf:\implies\dfrac{ 2^m -2^n}{2^5} = 3 \\\\\sf:\implies \dfrac{2^m}{2^5} -\dfrac{2^n}{2^5} =3 \\\\\sf:\implies 2^{(m-5)} -2^{(n-5)} = 4 -1 \qquad\bigg\lgroup \red{\tt Using \ \dfrac{a^k}{a^l}= a^{k-l}}\bigg\rgroup \\\\\sf:\implies 2^{(m-5)} -2^{(n-5)} = 2^2-2^0 \\\\\sf\qquad\qquad\tiny{ \underline\dag \:\:\sf \red{On \ Comparing \ respective \ terms }} \\\\\sf:\implies 2^{(m-5)}= 2^2 \qquad and \qquad 2^{(n-5)}= 2^0\\\\\sf:\implies m - 5=2\qquad and \qquad n-5 = 0\\\\\sf:\implies m = 5 +2 \qquad and \qquad n = 0 +5 \\\\:\implies\underset{\blue{\sf Required \ values}}{\underbrace{\boxed{\pink{\frak{ m = 7 \:\: , \:\: n = 5 }}}}}$

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H E N C E :-

• $\sf \orange{\tt n(A)+n(B) = 7+5=\purple{12} . }$
• $\sf \orange{\tt n(A)-n(B) = 7-5=\purple{2} . }$