Let S = {1, 2, 3, 4, 5, 6}. Then the number of one-one functions f: S → P(S), where P(S) denote the

Q: Wait, why do we have to choose 6 sizes out of 7? Shouldn't it just be 6?

Great question! The set $S$ has 6 elements, so its power set $P(S)$ has $2^6 = 64$ elements, with sizes ranging from 0 to 6. We are defining a function $f$ that maps the 6 elements of $S$ to 6 distinct subsets in $P(S)$. Because the condition $f(1) \subset f(2) \subset \dots \subset f(6)$ forces the sizes to be strictly increasing, we are essentially picking 6 distinct values from the 7 possible sizes $\{0, 1, 2, 3, 4, 5, 6\}$. Choosing 6 out of 7 is equivalent to deciding which one size to exclude.

Q: Why does the one-to-one condition imply strict inclusion?

The one-to-one condition means that $f(n) \neq f(m)$ for $n \neq m$. If we have a chain $f(1) \subset f(2) \subset \dots \subset f(6)$, and we know all these sets must be distinct, then it is impossible for any two sets in the chain to have the same number of elements. If $|f(n)| = |f(m)|$ for $n < m$, then $f(n)$ would have to be equal to $f(m)$, which violates the one-to-one condition. Therefore, the sizes must be strictly increasing.

Q: How do we get the product of combinations formula?

Think of it as a construction process. We start with the largest set $f(6)$, which has $s_6$ elements chosen from the 6 elements of $S$ ($\binom{6}{s_6}$ ways). Then, $f(5)$ must be a subset of $f(6)$ with $s_5$ elements, so we choose $s_5$ elements from the $s_6$ elements already in $f(6)$ ($\binom{s_6}{s_5}$ ways). We repeat this logic down the chain. The total number of ways is the product of these choices.

Question

Let $S = \{1, 2, 3, 4, 5, 6\}$. Then the number of one-one functions $f: S 	o P(S)$, where $P(S)$ denote the power set of S, such that $f(n) \subset f(m)$ where $n < m$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a set $S = \{1, 2, 3, 4, 5, 6\}$. This is our universe, a collection of six distinct entities.

We are tasked with creating a function $f$ that maps each element of $S$ to a subset of $S$, which is an element of the power set $P(S)$. This mapping is governed by the strict law of nested inclusion: $f(n) \subset f(m)$ whenever $n < m$.

### The Nested Doll Analogy

Think of the condition $f(1) \subset f(2) \subset f(3) \subset f(4) \subset f(5) \subset f(6)$ as a set of Russian nesting dolls. Each doll must fit perfectly inside the next.

Because the function is one-to-one, no two dolls can be the same size. If $f(1)$ were the same size as $f(2)$, they would have to be the same set, which would break our one-to-one requirement.

Thus, the sizes of these sets must be strictly increasing. Let $s_i = |f(i)|$. We have:
$$0 \le s_1 < s_2 < s_3 < s_4 < s_5 < s_6 \le 6$$

We are choosing six distinct sizes from the seven available integers $\{0, 1, 2, 3, 4, 5, 6\}$. The number of ways to choose these six sizes is simply $\binom{7}{6} = 7$.

### The Combinatorial Engine

Once we have fixed a sequence of sizes, we work from the outside in to count the number of ways to build these sets.

To form $f(6)$, we choose $s_6$ elements from the 6 available in $S$ in $\binom{6}{s_6}$ ways. To form $f(5)$, we must choose $s_5$ elements from the $s_6$ elements already chosen for $f(6)$, which can be done in $\binom{s_6}{s_5}$ ways.

Continuing this logic, the total number of ways for a fixed sequence of sizes is the product:
$$\binom{6}{s_6} 	imes \binom{s_6}{s_5} 	imes \binom{s_5}{s_4} 	imes \binom{s_4}{s_3} 	imes \binom{s_3}{s_2} 	imes \binom{s_2}{s_1}$$

Note that $s_1$ is the size of the smallest set, and we implicitly have $\binom{s_1}{0} = 1$ if $s_1=0$.

### Exploring the Cases

Let us examine the two extreme cases. If we omit the size 6, our sizes are $\{0, 1, 2, 3, 4, 5\}$. The number of ways is:
$$\binom{6}{5} 	imes \binom{5}{4} 	imes \binom{4}{3} 	imes \binom{3}{2} 	imes \binom{2}{1} 	imes \binom{1}{0} = 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 720$$

If we omit the size 0, our sizes are $\{1, 2, 3, 4, 5, 6\}$. The number of ways is:
$$\binom{6}{6} 	imes \binom{6}{5} 	imes \binom{5}{4} 	imes \binom{4}{3} 	imes \binom{3}{2} 	imes \binom{2}{1} = 1 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 = 720$$

Now, consider omitting a middle size, such as 5. Our sizes are $\{0, 1, 2, 3, 4, 6\}$. The calculation becomes:
$$\binom{6}{6} 	imes \binom{6}{4} 	imes \binom{4}{3} 	imes \binom{3}{2} 	imes \binom{2}{1} 	imes \binom{1}{0} = 1 	imes 15 	imes 4 	imes 3 	imes 2 	imes 1 = 360$$

This 'gap' of 2 between sizes 4 and 6 introduces a factor of 15, changing the result. This symmetry holds for all five intermediate cases (omitting 1, 2, 3, 4, or 5), each yielding 360.

### The Grand Total

We have two cases yielding 720 and five cases yielding 360. The total number of one-to-one functions is:
$$2 	imes 720 + 5 	imes 360 = 1440 + 1800 = 3240$$

The final answer is **3240**.

Animated Solution for Mathematics - Permutations and Combinations: Let $S = {1, 2, 3, 4, 5, 6}$ . Then the number of one-one functions $f : S \to P (S)$ , where $P (S)$ denote the power set of S, such that $f (n) \subset f (m)$ where $n < m$ is

Enter Numerical Value:

Visualized Solution

Understanding the Domain and Codomain

The Inclusion Condition

Strictly Increasing Sizes

Choosing Size Sequences

General Counting Formula

Case 1: Sizes ${0, 1, 2, 3, 4, 5}$

Case 7: Sizes ${1, 2, 3, 4, 5, 6}$

Case 2: Sizes ${0, 1, 2, 3, 4, 6}$

Symmetry of Intermediate Cases

Final Summation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Nested Doll Analogy

The Combinatorial Engine

Exploring the Cases

The Grand Total

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