Two finite sets have $m$ and $n$ elements. The number of subsets of the first set is 112 more than that of the second set. The values of $m$ and $n$ are, respectively, (A) $4,\ 7$ (B) $7,\ 4$ (C) $4,\ 4$ (D) $7,\ 7$

> [!TLDR] > The number of subsets of a set with $k$ elements is $2^k$. Setting $2^m = 2^n + 112$ and testing small powers of 2 gives $m = 7$, $n = 4$, so the correct option is **(B)**. The core idea here is simple: the number of subsets of a finite set depends only on its cardinality. If a set has $k$ elements, it has $2^k$ subsets (including the empty set). The problem gives a relationship between the subset counts of two sets, so we translate that into an equation in powers of 2. Let the first set have $m$ elements and the second $n$ elements. Then: - Number of subsets of the first set = $2^m$ - Number of subsets of the second set = $2^n$ The problem states: $2^m = 2^n + 112$. We need integer values of $m$ and $n$ (both positive, since they are sizes of finite sets). This is a Diophantine equation in powers of 2. The trick is to notice that $2^m$ and $2^n$ are powers of 2, and 112 is not a power of 2, so $m$ and $n$ cannot be equal (otherwise the difference would be 0). Also, $m > n$ because $2^m$ is larger. Let’s solve step by step. 1. **Rewrite the equation** $2^m - 2^n = 112$ Factor out the smaller power: $2^n (2^{m-n} - 1) = 112$. 2. **Factor 112 into a power of 2 times an odd number** $112 = 16 \times 7 = 2^4 \times 7$. So $2^n (2^{m-n} - 1) = 2^4 \times 7$. 3. **Match the factors** Since $2^n$ is a pure power of 2 and $(2^{m-n} - 1)$ is odd (any power of 2 minus 1 is odd), we must have: …

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Concept understanding — Subset Listing

Subset Listing: A First Look

Let's build this from the ground up — no jargon, just intuition first.

1. The Intuition: What does "subset" mean?

Imagine you have a set — a collection of distinct objects. For example:

Set A = {apple, banana, cherry}

Now, a subset is simply a selection of some (or all, or none) of these objects, taken from the original set.

You could pick all three → {apple, banana, cherry}
You could pick just two → {apple, banana}
You could pick just one → {cherry}
You could pick none → {} (the empty set)

Each of these is a subset of the original set.

2. The Precise Definition

Definition: A set $B$ is a subset of a set $A$ if every element of $B$ is also an element of $A$ .

We write this as:

B \subseteq A

If $B$ is not a subset of $A$ , we write:

B \not\subseteq A

Key points to remember:

Every set is a subset of itself.

Example: {apple, banana} ⊆ {apple, banana}
The empty set $\emptyset$ (or $\{\}$ ) is a subset of every set.

Why? Because it has no elements, so there's nothing to violate the condition.
If $B$ is a subset of $A$ but $B \neq A$ , we call $B$ a proper subset.

Notation: $B \subset A$ (some books use $\subsetneq$ )

3. How to "list" all subsets

Subset listing means writing down every possible subset of a given set.

Example: Set $S = \{a, b\}$

All subsets:

$\emptyset$ (empty set)
$\{a\}$
$\{b\}$
$\{a, b\}$ (the set itself)

So the list of all subsets is:

\{ \emptyset, \{a\}, \{b\}, \{a, b\} \}

How many subsets does a set have?

If a set has $n$ elements, it has exactly $2^n$ subsets.

$n = 0$ → $2^0 = 1$ subset (just the empty set)
$n = 1$ → $2^1 = 2$ subsets
$n = 2$ → $2^2 = 4$ subsets (as above)
$n = 3$ → $2^3 = 8$ subsets

Why $2^n$ ?

For each element, you have 2 choices: include it or exclude it. Multiply these choices: $2 \times 2 \times \dots \times 2$ ( $n$ times) = $2^n$ .

4. A systematic way to list subsets

For a set with $n$ elements, you can use a binary counting method:

Label each element with a position (1st, 2nd, 3rd, ...)
Count from $0$ to $2^n - 1$ in binary
Each binary number tells you which elements to include (1 = include, 0 = exclude)

Example: $S = \{a, b, c\}$ (3 elements)

Binary	Subset
000	$\emptyset$
001	$\{c\}$
010	$\{b\}$
011	$\{b, c\}$
100	$\{a\}$

Why this formula?

Okay, let's break down Subset Listing from the ground up. The core idea is simple: given a set, how do we systematically list all its subsets, and why does the formula $2^n$ work?

1. The Core Question

Imagine you have a set with $n$ elements, like $S = \{a, b, c\}$ (so $n=3$ ). A subset is any collection of elements from $S$ , including the empty set $\{\}$ and the set itself $\{a, b, c\}$ .

The key formula is:

Total number of subsets of a set with $n$ elements = $\mathbf{2^n}$

Let's see why this is true, not just memorize it.

2. The "Decision" or "Binary Choice" Reasoning

The most intuitive derivation comes from thinking about each element individually.

For each element in the original set, when building a subset, you have exactly two choices:

Include the element in the subset.
Exclude the element from the subset.

This is a fundamental, independent decision for every element.

Example with $S = \{a, b, c\}$

For element $a$ : Choose IN or OUT. (2 choices)
For element $b$ : Choose IN or OUT. (2 choices)
For element $c$ : Choose IN or OUT. (2 choices)

Since these choices are independent (choosing for $a$ doesn't affect the choice for $b$ ), the total number of distinct combinations of choices is the product of the number of choices for each element:

2 \times 2 \times 2 = 2^3 = 8

This directly gives the 8 subsets of $\{a, b, c\}$ :

$\{\}$ (all OUT)
$\{a\}$ (a IN, b OUT, c OUT)
$\{b\}$
$\{c\}$
$\{a, b\}$
$\{a, c\}$
$\{b, c\}$
$\{a, b, c\}$ (all IN)

3. The General Formula (Derivation)

For a set with $n$ elements, you have $n$ independent binary decisions. Therefore:

\text{Total subsets} = \underbrace{2 \times 2 \times \dots \times 2}_{n \text{ times}} = 2^n

This is the fundamental reason the formula holds. It's not a coincidence; it's a direct consequence of the counting principle for independent events.

4. Why This Matters for Exams

Don't just memorize $2^n$ . If a question asks "How many subsets does a set with 5 elements have?", you can instantly say $2^5 = 32$ . But if they ask why, you now have the reasoning. …

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The number of subsets of a set with $k$ elements is $2^k$ . Setting $2^m = 2^n + 112$ and testing small powers of 2 gives $m = 7$ , $n = 4$ , so the correct option is (B).

The core idea here is simple: the number of subsets of a finite set depends only on its cardinality. If a set has $k$ elements, it has $2^k$ subsets (including the empty set). The problem gives a relationship between the subset counts of two sets, so we translate that into an equation in powers of 2.

Let the first set have $m$ elements and the second $n$ elements. Then:

Number of subsets of the first set = $2^m$
Number of subsets of the second set = $2^n$

The problem states: $2^m = 2^n + 112$ .

We need integer values of $m$ and $n$ (both positive, since they are sizes of finite sets). This is a Diophantine equation in powers of 2. The trick is to notice that $2^m$ and $2^n$ are powers of 2, and 112 is not a power of 2, so $m$ and $n$ cannot be equal (otherwise the difference would be 0). Also, $m > n$ because $2^m$ is larger.

Let’s solve step by step.

Rewrite the equation

$2^m - 2^n = 112$

Factor out the smaller power: $2^n (2^{m-n} - 1) = 112$ .
Factor 112 into a power of 2 times an odd number

$112 = 16 \times 7 = 2^4 \times 7$ .

So $2^n (2^{m-n} - 1) = 2^4 \times 7$ .
Match the factors

Since $2^n$ is a pure power of 2 and $(2^{m-n} - 1)$ is odd (any power of 2 minus 1 is odd), we must have: …

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Q.Two finite sets have mmm and nnn elements. The number of subsets of the first set is 112 more than that of the second set. The values of mmm and nnn are, respectively, (A) 4, 74,\ 74, 7 (B) 7, 47,\ 47, 4 (C) 4, 44,\ 44, 4 (D) 7, 77,\ 77, 7

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Q.Two finite sets have $m$ and $n$ elements. The number of subsets of the first set is 112 more than that of the second set. The values of $m$ and $n$ are, respectively,
(A) $4,\ 7$
(B) $7,\ 4$
(C) $4,\ 4$
(D) $7,\ 7$

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