`p` is a prime number and `nltplt2n dot` If `N=^(2n)C_n ,` then

Options

A. p divides N
B. `p^(2)` divides N
C. p cannot divides N
D. None of these

Answer

To solve the problem, we need to analyze the expression given and the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Problem**: We are given that \( p \) is a prime number and \( n < p < 2n \). We need to find a relationship involving \( N = \binom{2n}{n} \).

2. **Using the Binomial Coefficient**: The binomial coefficient \( \binom{2n}{n} \) can be expressed as:
\[ N = \binom{2n}{n} = \frac{(2n)!}{(n!)^2} \]

3. **Expanding the Factorial**: The factorial \( (2n)! \) can be expanded as:
\[ (2n)! = 1 \cdot 2 \cdot 3 \cdots (2n) \]
and \( n! \) can be expanded as:
\[ n! = 1 \cdot 2 \cdot 3 \cdots n \]

4. **Simplifying the Expression**: Therefore, we can rewrite \( N \) as:
\[ N = \frac{1 \cdot 2 \cdot 3 \cdots (2n)}{(1 \cdot 2 \cdots n)(1 \cdot 2 \cdots n)} = \frac{(2n)(2n-1)(2n-2)\cdots(n+1)}{n!} \]
Here, the numerator consists of the terms from \( n+1 \) to \( 2n \).

5. **Analyzing the Prime \( p \)**: Since \( p \) is a prime number and it lies between \( n \) and \( 2n \), it will appear exactly once in the numerator of the expression for \( N \) but not in the denominator.

6. **Conclusion about Divisibility**: Since \( p \) appears in the numerator and not in the denominator, it follows that \( p \) must divide \( N \). Hence, we conclude that:
\[ p \text{ divides } N \]

### Final Result:
The answer is that \( p \) divides \( N \).