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(B) The number of subsets of a set with $k$ elements is $2^k$. Given that the first set has $m$ elements and the second set has $n$ elements,the numbe

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$6, 3$

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(B) The number of subsets of a set with $k$ elements is $2^k$. Given that the first set has $m$ elements and the second set has $n$ elements,the number of subsets are $2^m$ and $2^n$ respectively. According to the problem,$2^m - 2^n = 56$. $2^n(2^{m-n} - 1) = 56$. We can write $56$ as $8 	imes 7 = 2^3 	imes (8 - 1) = 2^3 	imes (2^3 - 1)$. Comparing $2^n(2^{m-n} - 1) = 2^3(2^3 - 1)$,we get $n = 3$ and $m - n = 3$. Substituting $n = 3$ into $m - n = 3$,we get $m - 3 = 3$,so $m = 6$. Thus,the values are $m = 6$ and $n = 3$.

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

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