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Question: Two finite sets have m and n elements. The total number of subsets of the first set is 48 more than ...

Two finite sets have m and n elements. The total number of subsets of the first set is 48 more than the total number of subsets of the second set. The values of m and n are?
A. 7,6 B. 6,7 C. 6,4 D. 7,4  {\text{A}}{\text{. 7,6}} \\\ {\text{B}}{\text{. 6,7}} \\\ {\text{C}}{\text{. 6,4}} \\\ {\text{D}}{\text{. 7,4}} \\\

Explanation

Solution

Hint- No. of subsets of a finite set of xxelements is 2x{2^x}.We have to find the no of subsets of both the finite sets of given no of elements and form an equation that satisfies the given statement.

Complete step-by-step solution -
No. of subsets for first set =2m = {2^m}
No. of subsets for second set =2n = {2^n}
Now, according to question,
No. of subsets of first set is 48 more than that of second set

2m=48+2n 2m2n=48 Now factorize 48 and take 2n common from left side 2n(2m2n1)=16×3 2n(2mn1)=24×(41) 2n(2mn1)=24×(221)  {2^m} = 48 + {2^n} \\\ {2^m} - {2^n} = 48 \\\ {\text{Now factorize 48 and take }}{{\text{2}}^n}{\text{ common from left side}} \\\ {2^n}(\dfrac{{{2^m}}}{{{2^n}}} - 1) = 16 \times 3 \\\ {2^n}({2^{m - n}} - 1) = {2^4} \times (4 - 1) \\\ {2^n}({2^{m - n}} - 1) = {2^4} \times ({2^2} - 1) \\\

Here we have factorized 48 in such a way that we can get the value of m and n by comparing LHS and RHS.
Comparing both sides, we get
n=4 and mn=2 m4=2 m=4+2 m=6  n = 4 \\\ {\text{and }}m - n = 2 \\\ m - 4 = 2 \\\ m = 4 + 2 \\\ m = 6 \\\
Therefore, the correct option is C.

Note- For solving such a question try to consider the number of elements in some unknown variable and with the help of the problem statement try to bring out some algebraic equation. The formula for number of subsets for any given set must be remembered.