The product of r consecutive integers is divisible by

Options

A. `r!`
B. `(r-1)!`
C. `(r+1)!`
D. none of these

Answer

To solve the problem of determining the divisibility of the product of \( r \) consecutive integers, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Product of Consecutive Integers**:
The product of \( r \) consecutive integers can be represented as:
\[ n \times (n-1) \times (n-2) \times \ldots \times (n-r+1) \]
where \( n \) is the largest integer in the sequence.

2. **Using Factorial Representation**:
We can express the product of these \( r \) consecutive integers using factorials. The product can be written as:
\[ \frac{n!}{(n-r)!} \]
This is because \( n! \) is the product of all integers from \( 1 \) to \( n \), and dividing by \( (n-r)! \) removes the product of integers from \( 1 \) to \( n-r \).

3. **Divisibility by \( r! \)**:
We need to show that the product of these \( r \) consecutive integers is divisible by \( r! \). The factorial \( r! \) is the product of the first \( r \) integers:
\[ r! = r \times (r-1) \times (r-2) \times \ldots \times 1 \]

4. **Applying the Concept of Combinations**:
The number of ways to choose \( r \) items from \( n \) items is given by the binomial coefficient:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
This implies that \( n! \) can be expressed as:
\[ n! = r! \cdot \binom{n}{r} \cdot (n-r)! \]
Therefore, \( n! \) is divisible by \( r! \).

5. **Conclusion**:
Since \( \frac{n!}{(n-r)!} \) represents the product of \( r \) consecutive integers, and we have established that \( n! \) is divisible by \( r! \), it follows that the product of \( r \) consecutive integers is also divisible by \( r! \).

Thus, we conclude that the product of \( r \) consecutive integers is divisible by \( r! \).

### Final Answer:
The product of \( r \) consecutive integers is divisible by \( r! \).