The total number of injective mappings from a set with m elements to a set with n elements, `m <= n`,is

Options

A. `m^(n)`
B. `n^(m)`
C. `(n!)/((n-m)!)`
D. `n^(!)`

Answer

To find the total number of injective mappings (or one-to-one functions) from a set with \( m \) elements to a set with \( n \) elements, where \( m \leq n \), we can use the following reasoning:

### Step-by-Step Solution:

1. **Understanding Injective Mappings**: An injective mapping from set A (with \( m \) elements) to set B (with \( n \) elements) means that each element in A maps to a unique element in B. No two elements in A can map to the same element in B.

2. **Choosing the First Element**: For the first element of the set with \( m \) elements, we have \( n \) choices in the set with \( n \) elements.

3. **Choosing the Second Element**: For the second element of the set with \( m \) elements, we can only choose from the remaining \( n - 1 \) elements in the set with \( n \) elements (since the first element has already been chosen).

4. **Continuing the Process**: This process continues for all \( m \) elements. For the third element, we have \( n - 2 \) choices, and so on.

5. **Generalizing the Choices**: Therefore, the total number of choices can be expressed as:
\[ n \times (n - 1) \times (n - 2) \times \ldots \times (n - m + 1) \]

6. **Using Factorials**: This product can be rewritten using factorials:
\[ = \frac{n!}{(n - m)!} \]
Here, \( n! \) is the factorial of \( n \), and \( (n - m)! \) is the factorial of \( n - m \).

7. **Final Result**: Thus, the total number of injective mappings from a set with \( m \) elements to a set with \( n \) elements is given by:
\[ \frac{n!}{(n - m)!} \]

### Conclusion:
The total number of injective mappings from a set with \( m \) elements to a set with \( n \) elements, where \( m \leq n \), is:
\[ \frac{n!}{(n - m)!} \]