The product of `n` consecutive natural numbers is always divisible by

Options

A. (A)divisible by n!
B. (B)divisible by (n+1)!
C. (C)odd
D. (D)Multiple of 4

Explanation

To solve the problem of determining what the product of `n` consecutive natural numbers is always divisible by, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Consecutive Natural Numbers**:
Let the `n` consecutive natural numbers be represented as:
\[ k + 1, k + 2, k + 3, \ldots, k + n \]
where \( k \) is a non-negative integer (i.e., \( k \geq 0 \)).

2. **Express the Product**:
The product of these `n` consecutive natural numbers can be expressed as:
\[ P = (k + 1)(k + 2)(k + 3) \ldots (k + n) \]

3. **Factorial Representation**:
We can relate this product to factorials. The product \( P \) can be rewritten using factorials:
\[ P = \frac{(k + n)!}{k!} \]
This is because the factorial \( (k + n)! \) includes all the numbers from 1 to \( k + n \), and dividing by \( k! \) removes the first \( k \) terms.

4. **Divisibility by \( n! \)**:
To show that \( P \) is divisible by \( n! \), we can multiply and divide by \( n! \):
\[ P = \frac{(k + n)!}{k!} = \frac{(k + n)!}{n!} \cdot n! \]
The term \( \frac{(k + n)!}{n!} \) represents the number of ways to choose \( n \) items from \( k + n \) items, which is a binomial coefficient:
\[ P = \binom{k + n}{n} \cdot n! \]

5. **Conclusion**:
Since \( P \) can be expressed as \( \binom{k + n}{n} \cdot n! \), it is clear that \( P \) is always divisible by \( n! \). Therefore, we conclude that the product of `n` consecutive natural numbers is always divisible by \( n! \).

### Final Answer:
The product of `n` consecutive natural numbers is always divisible by \( n! \).