If `n lt p lt 2n` and p is prime and `N=.^(2n)C_(n)`, then

Options

A. n divides N
B. p does not divide N
C. `p^(2)` divides N
D. `p^(2)` does not divide N

Answer

To solve the problem, we need to analyze the given conditions step by step.

### Step 1: Understand the given conditions
We have the conditions:
1. \( n < p < 2n \)
2. \( p \) is a prime number.
3. \( N = \binom{2n}{n} \)

### Step 2: Write the expression for \( N \)
The binomial coefficient \( \binom{2n}{n} \) can be expressed as:
\[ N = \frac{(2n)!}{(n!)^2} \]

### Step 3: Factorial representation
We can expand \( (2n)! \) and \( n! \):
\[ N = \frac{(2n)(2n-1)(2n-2)\ldots(n+1)}{n!} \]
This is because \( (2n)! \) can be expressed as \( (2n)(2n-1)(2n-2)\ldots(n+1) \times n! \).

### Step 4: Simplifying the expression
The expression simplifies to:
\[ N = \frac{(2n)(2n-1)(2n-2)\ldots(n+1)}{n!} \]
This means we are multiplying \( n \) terms from \( (2n)(2n-1)\ldots(n+1) \).

### Step 5: Analyze divisibility by \( p \)
Since \( p \) is a prime number and \( n < p < 2n \), we need to check if \( N \) is divisible by \( p \).

### Step 6: Count the multiples of \( p \)
In the product \( (n+1)(n+2)\ldots(2n) \), we need to see how many of these terms are divisible by \( p \):
- The multiples of \( p \) in this range are \( p, 2p, \ldots \) up to \( 2n \).
- Since \( p < 2n \), there will be at least one multiple of \( p \) in the range \( (n+1) \) to \( (2n) \).

### Step 7: Conclusion about \( N \)
Since \( N \) includes a factor of \( p \) from the terms \( (n+1) \) to \( (2n) \), we conclude that \( N \) is divisible by \( p \).

### Step 8: Finding the value of \( n \)
Given that \( p \) is prime and \( n < p < 2n \), we can deduce that \( n \) must be less than \( p \) and \( p \) must be greater than \( n \). The only way this holds true is if \( n \) is a prime number itself.

### Final Answer
Thus, the value of \( n \) must be a prime number that satisfies the conditions \( n < p < 2n \).

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