Let A and B be two finite sets having m and n elements respectively such that m≤n . A mapping is selected at random from the set of all mappings from A to B. The probability that the mapping selected is an injection, is

In above question if `m gt n` then the probability that the mapping selected is an injective mapping is :

Options

A. `(n !)/( (n - m) !m^(n))`
B. `(n!)/((n - m) !n^(m))`
C. `(""^(n)C_(m))/(n^(m))`
D. None of these

Answer

To solve the problem, we will analyze the two cases for the sets A and B, which have m and n elements respectively. We will derive the probability of selecting an injective mapping from A to B.

### Case 1: When \( m \leq n \)

1. **Understanding the Problem**: An injective mapping (or one-to-one function) means that each element in set A maps to a unique element in set B. No two elements in A can map to the same element in B.

2. **Counting Favorable Outcomes**:
- The first element of A can map to any of the n elements in B.
- The second element can map to any of the remaining \( n - 1 \) elements.
- The third element can map to \( n - 2 \), and so on.
- Thus, the total number of favorable outcomes for an injective mapping is:
\[ n \times (n - 1) \times (n - 2) \times \ldots \times (n - m + 1 = \frac{n!}{(n - m)!} \]

3. **Counting Total Outcomes**:
- Each of the m elements in A can map to any of the n elements in B independently.
- Therefore, the total number of mappings from A to B is:
\[ n^m \]

4. **Calculating Probability**:
- The probability \( P(E) \) that a randomly selected mapping is injective is given by:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{\frac{n!}{(n - m)!}}{n^m} \]

### Case 2: When \( m > n \)

1. **Understanding the Problem**: If \( m > n \), it is impossible to have an injective mapping because there are more elements in A than in B. Thus, at least one element in A would have to map to the same element in B.

2. **Counting Favorable Outcomes**:
- Since an injective mapping is not possible, the number of favorable outcomes is 0.

3. **Counting Total Outcomes**:
- The total number of mappings from A to B remains \( n^m \).

4. **Calculating Probability**:
- The probability \( P(E) \) that a randomly selected mapping is injective is:
\[ P(E) = \frac{0}{n^m} = 0 \]

### Final Answers:
- For \( m \leq n \):
\[ P(E) = \frac{n!}{(n - m)! \cdot n^m} \]
- For \( m > n \):
\[ P(E) = 0 \]