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Exercise 3.3 · Q5

Q.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. Find the values of 'm' and 'n'.

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The number of subsets of a set with kk elements is 2k2^k; setting up 2m=2n+562^m = 2^n + 56 and factoring pins down mm and nn uniquely.

Number of subsets of a set with kk elements: 2k2^k.

  1. Translate the statement. Let the first set have mm elements and the second have nn elements. Subsets of the first =2m=2^m, subsets of the second =2n=2^n. Given: 2m=2n+562^m = 2^n + 56, i.e. 2m−2n=562^m - 2^n = 56.

  2. Factor out 2n2^n (taking m>nm>n): 2n(2m−n−1)=562^n(2^{m-n} - 1) = 56.

  3. Factorise 56. 56=23×7=8×756 = 2^3 \times 7 = 8 \times 7. Since 2m−n−12^{m-n}-1 is always odd (one less than a power of 2, for m>nm>n) and 77 is odd while 8=238=2^3 is the power-of-2 factor, match: 2n=82^n = 8 and 2m−n−1=72^{m-n}-1 = 7. …

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