Statement-1: Let A and B be two sets having m and n elements respectively such that `m lt n`. Then,
Number of surjections from A to B `=sum_(r=1)^(n) ""^(n)C_(r) (-1)^(n-r) r^(m)`
Statement-2: If `f:A to B` is a surjection, then every element in B has a pre-image in A.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze both statements provided in the question regarding surjective functions and the relationship between sets A and B.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- A **surjection** (or onto function) from set A to set B is a function where every element in B has at least one pre-image in A.
- If set A has \( m \) elements and set B has \( n \) elements, and if \( m < n \), it implies that there are more elements in B than in A.

2. **Analyzing Statement 2**:
- Statement 2 claims: "If \( f: A \to B \) is a surjection, then every element in B has a pre-image in A."
- By definition of a surjective function, this statement is true. A surjective function ensures that every element in B is mapped from at least one element in A.

3. **Analyzing Statement 1**:
- Statement 1 claims: "The number of surjections from A to B is given by the formula \( \sum_{r=1}^{n} \binom{n}{r} (-1)^{n-r} r^{m} \)."
- Given that \( m < n \), we need to consider the implications of this condition. If A has fewer elements than B, it is impossible to create a surjective function from A to B because there are not enough elements in A to cover all elements in B.

4. **Conclusion for Statement 1**:
- Since there cannot be any surjective functions from A to B when \( m < n \), the number of surjections is indeed 0. Thus, the formula provided in Statement 1 does not hold true in this case, making Statement 1 false.

5. **Final Evaluation**:
- Statement 1 is false (no surjective functions exist when \( m < n \)).
- Statement 2 is true (a surjective function by definition maps every element in B).

### Final Answer:
- Statement 1: False
- Statement 2: True