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Question: Two finite sets have elements. The total number of subsets of the first set is 48 more than the tota...

Two finite sets have elements. The total number of subsets of the first set is 48 more than the total number of subsets of the second set. Find m and nm \text{ and } n .

Explanation

Solution

In order to answer this question, to know the value of m and nm \text{ and } n with respect to the given question, first we will assume two variables which have m and nm \text{ and } n number of elements. Then we will follow the instructions given in the question itself.

Complete step-by-step solution:
Let A have mm elements.
Let B have nn elements.
Total number of subsets of A =2m = {2^m}
Total number of subsets of B =2n = {2^n}
According to the question, now we have an equation:-
2m2n=48{2^m} - {2^n} = 48
Now, we will take common 2n{2^n} from the L.H.S-
2n(2mn2)=48\Rightarrow {2^n}({2^{m - n}} - 2) = 48
So, 2n=even and 2mn2=0even{2^n} = \text{even}\,\text{ and }\,{2^{m - n}} - 2 = 0\,even
Now,
48=8×6=23×61 2n(2mn2)=23×6 n=3 \begin{aligned} &48 = 8 \times 6 = {2^3} \times {6^1} \\\ & \Rightarrow {2^n}({2^{m - n}} - 2) = {2^3} \times 6 \\\ &\Rightarrow n = 3 \\\ \end{aligned}
Now,
8(2m32)=6×8 2m32=6 2m3=8 2m3=23 m3=3 m=6 \begin{aligned} &8({2^{m - 3}} - 2) = 6\times 8 \\\ &\Rightarrow {2^{m - 3}} - 2 = 6 \\\ & \Rightarrow {2^{m - 3}} = 8 \\\ & \Rightarrow {2^{m - 3}} = {2^3} \\\ &\Rightarrow m - 3 = 3 \\\ & \Rightarrow m = 6 \\\ \end{aligned}
Therefore, the value of m and nm \text{ and } n are 6 and 3 respectively.

Note: A subset is a collection of elements that also appear in another collection. Remember that a set is a group of related elements. For example, a,b,c,d\\{ a,b,c,d\\} is a set of letters, while cat,dog,fish,bird\\{ cat,dog,fish,bird\\} is a set of animals. 2,4,6,8,10\\{ 2,4,6,8,10\\} is a set of even numbers, and a,b,c,d\\{ a,b,c,d\\} is a set of letters.