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1.Set Theory
medium
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
A
$7, 6$
B
$6, 3$
C
$5, 1$
D
$8, 7$
Solution
(b) Since ${2^m} – {2^n} = 56 = 8 \times 7 = {2^3} \times 7$
==> ${2^n}({2^{m – n}} – 1) = {2^3} \times 7$, $\therefore $ $n = 3$ and ${2^{m – n}} = 8 = {2^3}$
==> $m – n = 3$ ==> $m – 3 = 3$ ==> $m = 6$; $\therefore \,\,m = 6,\,\,n = 3$.
Standard 11
Mathematics
