Question & Answer
QUESTION

Two finite sets have m and n elements. The number of elements in the power set of the first is 48 more than the total number of elements in the power set of the second set. Find m and n.

ANSWER Verified Verified
Hint: First of all, try to recollect what power set is, the number of elements in it. Consider the given two sets as Set A and Set B. Now write the number of elements in the power set of Set A and Set B and equate their difference to 48. From here estimate the values of m and n.

Complete step-by-step answer:
Here we are given that two finite sets have m and n elements. The number of elements in the power set of first is 48 more than the total number of elements in the power set of the second set. We have to find the values of m and n. Before proceeding with the question, let us first talk about a few terms.

1. Sets: Sets are represented as a collection of well-defined objects or elements and it does not change from person to person. A set is represented by a capital letter symbol and the number of elements in the finite set is represented as the cardinal number of a set. For example, A = {1, 2, 3, 4} is a set.

2. Subsets: Subsets are a part of the sets. For example, if set A is a collection of even numbers and set B consists of {2, 4, 6}, then B is said to be a subset of A, denoted by \[B\subseteq A\].

3. Power Set: The power of a set A is defined as the set of all subsets of the Set A including the set itself and the null or empty set. It is denoted by P(A). This set is the combination of all subsets including the null set, of a given set. If the given set has elements, then its Power Set will contain \[{{2}^{a}}\] elements.

Now let us consider our question. Let us assume that the given two sets are Set A and Set B. Now we are given that these two sets have m and n elements. So, we get,
Number of elements in Set A = m
Number of elements in Set B = n
Now, we know that, if the given set has ‘a’ elements, then its Power Set will contain \[{{2}^{a}}\] elements. By using this, we get,
Number of elements in the power set of Set A \[={{2}^{m}}.....\left( i \right)\]
Number of elements in the power set of Set B \[={{2}^{n}}.....\left( ii \right)\]
Now we are given that the number of elements in the power set of the first set is 48 more than the number of elements in the power set of the second set. So, we get,
(Number of elements in the power set of Set A) – (Number of elements in the power set of Set B) = 48
By substituting the respective values from equation (i) and (ii) in the above equation, we get,
\[{{2}^{m}}-{{2}^{n}}=48\]
By taking out \[{{2}^{n}}\] common, we get,
\[{{2}^{n}}\left( {{2}^{m-n}}-1 \right)=48\]
We can also write the above equation as,
\[{{2}^{n}}\left( {{2}^{m-n}}-1 \right)=16\times 3\]
By substituting \[16={{2}^{4}}\] and \[3={{2}^{2}}-1\] in the above equation, we get,
\[{{2}^{n}}\left( {{2}^{m-n}}-1 \right)={{2}^{4}}\left( {{2}^{2}}-1 \right)\]
By comparing LHS and RHS of the above equation we get,
\[n=4\]
\[m-n=2\]
By substituting n= 4 in the above equation, we get,
\[m-4=2\]
\[m=2+4=6\]
Hence, we get the value of m and n as 6 and 4 respectively.

Note: In these types of questions, where we don’t have two equations to solve for two variables (here m and n) individually, we express RHS in terms of the power of bases according to LHS and then compare the expressions at LHS and RHS to get the value of various variables. The power at RHS can be estimated by hit and trial of different numbers.