Let `S={1,2,3,4,5,6}`.Then the number of one-one functions `f:S rarr P(S)`,where `P(S)` denote the power set of `S`,such that `f(n) sub f(m)` where `n lt m` is

Answer

To solve the problem of finding the number of one-one functions \( f: S \to P(S) \) such that \( f(n) \subset f(m) \) for \( n < m \), we will break it down step by step.

### Step 1: Understand the Sets
Let \( S = \{1, 2, 3, 4, 5, 6\} \) and \( P(S) \) be the power set of \( S \). The power set \( P(S) \) contains all subsets of \( S \). The number of elements in \( P(S) \) is \( 2^6 = 64 \).

### Step 2: Define the Function
We need to define a one-one function \( f \) from \( S \) to \( P(S) \) such that for any \( n < m \), \( f(n) \subset f(m) \). This means that the image of the function must be a chain of subsets where each subset is contained in the next.

### Step 3: Choose Subsets
To satisfy the condition \( f(n) \subset f(m) \), we can think of selecting \( k \) distinct subsets from \( P(S) \) such that they are ordered by inclusion. We can choose \( k \) subsets from \( P(S) \) in a way that they form a chain.

### Step 4: Count the Chains
The number of ways to choose \( k \) subsets from \( P(S) \) such that they are ordered by inclusion can be calculated using the binomial coefficients. For each \( k \), the number of ways to select \( k \) subsets from \( P(S) \) is given by \( \binom{n}{k} \), where \( n \) is the number of elements in \( S \).

### Step 5: Calculate for Each Case
We will consider cases for \( k = 1 \) to \( k = 6 \):
- For \( k = 1 \): Choose 1 subset from 64. There are \( \binom{64}{1} \) ways.
- For \( k = 2 \): Choose 2 subsets from 64. There are \( \binom{64}{2} \) ways.
- For \( k = 3 \): Choose 3 subsets from 64. There are \( \binom{64}{3} \) ways.
- For \( k = 4 \): Choose 4 subsets from 64. There are \( \binom{64}{4} \) ways.
- For \( k = 5 \): Choose 5 subsets from 64. There are \( \binom{64}{5} \) ways.
- For \( k = 6 \): Choose 6 subsets from 64. There are \( \binom{64}{6} \) ways.

### Step 6: Sum the Cases
The total number of one-one functions is given by the sum of all these cases:
\[ \text{Total} = \sum_{k=1}^{6} \binom{64}{k} \]

### Step 7: Calculate the Total
Using the binomial theorem, we know:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
Thus,
\[ \sum_{k=1}^{6} \binom{64}{k} = 2^{64} - 1 - \binom{64}{0} = 2^{64} - 1 - 1 = 2^{64} - 2 \]

### Final Answer
The total number of one-one functions \( f: S \to P(S) \) such that \( f(n) \subset f(m) \) for \( n < m \) is \( 2^{64} - 2 \).