The number of all one-one functions from the set [1,2,3,…,n] to itself is:

Options

A. n
B. `n^2`
C. `n^n`
D. n!

Answer

To find the number of all one-one functions from the set \([1, 2, 3, \ldots, n]\) to itself, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding One-One Functions**:
A one-one function (or injective function) from a set \(A\) to itself means that each element in set \(A\) must map to a unique element in set \(A\). No two elements in the domain can map to the same element in the codomain.

2. **Set Definition**:
Let the set \(A = \{1, 2, 3, \ldots, n\}\). We need to find the number of one-one functions from this set to itself.

3. **Choosing Mappings**:
- For the first element (1), we have \(n\) choices (it can map to any of the \(n\) elements).
- For the second element (2), we have \(n-1\) choices (since one element has already been used by the first element).
- For the third element (3), we have \(n-2\) choices (two elements have already been used).
- This pattern continues until the last element \(n\), which will have only 1 choice left.

4. **Calculating Total Choices**:
The total number of choices can be expressed as:
\[ n \times (n-1) \times (n-2) \times \ldots \times 1 \]
This product is known as \(n!\) (n factorial).

5. **Conclusion**:
Therefore, the number of all one-one functions from the set \([1, 2, 3, \ldots, n]\) to itself is:
\[ n! \]

### Final Answer:
The number of all one-one functions from the set \([1, 2, 3, \ldots, n]\) to itself is \(n!\).

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