First slide

Operations on Sets

Question

Two finite sets have m and n elements. 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 are

Easy

A

7,6
B

6,3
C

5,1
D

8,7

Solution

since,
$\begin{aligned} 2^{m} - 2^{n} = & 56 = 8 \times 7 = 2^{3} \times 7 \\ 2^{n} (2^{m - n} - 1) = 2^{3} \times 7 \end{aligned}$
$\begin{array}{ll} \Rightarrow n = 3 and 2^{m - n} = 8 = 2^{3} \Rightarrow n = 3 and m - n = 3 \\ \Rightarrow n = 3 and m - 3 = 3 \Rightarrow n = 3 and m = 6 \end{array}$

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