KCET · Maths · Sets and Relations
Two finite sets have \(m\) and \(n\) elements respectively. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of \(m\) and \(n\), respectively are
- A 7,6
- B 5,1
- C 6,3
- D 8,7
Answer & Solution
Correct Answer
(C) 6,3
Step-by-step Solution
Detailed explanation
Let set \(A\) have \(m\) elements and set \(B\) have \(n\) elements
Now, \(\quad 2^m-2^n=56\)
\(\Rightarrow \quad 2^n\left(2^{m-n}-1\right)=8 \times 7\)
\(\Rightarrow \quad 2^n\left(2^{m-n}-1\right)=2^3 \times 7\)
On comparing both sides, we get
\(2^n=2^3\) or \(2^{m-n}-1=7\)
Now, \(n=3\) ....(i)
So, \(\quad 2^{m-n}-1=7\)
\(\Rightarrow \quad 2^{m-n}=8\)
\(\Rightarrow \quad m-n=3\) ....(ii)
From Eqs. (i) and (ii), we get
\(m=6\) and \(n=3\)
Now, \(\quad 2^m-2^n=56\)
\(\Rightarrow \quad 2^n\left(2^{m-n}-1\right)=8 \times 7\)
\(\Rightarrow \quad 2^n\left(2^{m-n}-1\right)=2^3 \times 7\)
On comparing both sides, we get
\(2^n=2^3\) or \(2^{m-n}-1=7\)
Now, \(n=3\) ....(i)
So, \(\quad 2^{m-n}-1=7\)
\(\Rightarrow \quad 2^{m-n}=8\)
\(\Rightarrow \quad m-n=3\) ....(ii)
From Eqs. (i) and (ii), we get
\(m=6\) and \(n=3\)
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