Practicing Success
Practicing Success
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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 second set. The values of m and n are |
7, 6 6, 3 5, 1 8, 7 |
6, 3 |
$2^m–2^n= 56⇒2^n(2^{m-n}-1)=2^3×7$ by comparison $n=3$ $2^{m-3}-1=7⇒2^{m-3}=8$ $⇒m-3=3⇒m=6$ m = 6 and n = 3 |
