STATEMENT - 1 :
If `.^(2n+1)C_(1)` + `.^(2n+1)C_(2)+………+ .^(2n+1)C_(n)= 4095`, then `n = 7`
STATEMENT - 2 :
`.^(n)C_(r )= .^(n)C_(n-r)` where `n epsilon N, r epsilon W and n ge r`

Options

A. STATEMENT - 1 is True, STATEMENT- 2 is True , STATEMENT - 2 is a correct explanation for STATEMENT - 1
B. STATEMENT - 1 is True, STATEMENT - 2 is True , STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1
C. STATEMENT -1 is True, STATEMENT - 2 is False
D. STATEMENT -1 is False, STATEMENT - 2 is True

Explanation

To solve the problem, we need to analyze the statements given and verify their correctness step by step.

### Step 1: Analyze Statement 1
The statement is:
\[ \binom{2n+1}{1} + \binom{2n+1}{2} + \ldots + \binom{2n+1}{n} = 4095 \]

This expression represents the sum of the first \( n \) binomial coefficients from \( \binom{2n+1}{1} \) to \( \binom{2n+1}{n} \).

### Step 2: Use the Binomial Theorem
According to the binomial theorem, the sum of all coefficients in the expansion of \( (1 + 1)^{2n+1} \) is:
\[ 2^{2n+1} = \sum_{k=0}^{2n+1} \binom{2n+1}{k} \]
This means:
\[ \sum_{k=0}^{n} \binom{2n+1}{k} + \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} = 2^{2n+1} \]

### Step 3: Relate the Sums
Since \( \binom{2n+1}{k} = \binom{2n+1}{2n+1-k} \), we can say:
\[ \sum_{k=0}^{n} \binom{2n+1}{k} = \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} \]
Thus:
\[ 2 \sum_{k=0}^{n} \binom{2n+1}{k} = 2^{2n+1} \]
This simplifies to:
\[ \sum_{k=0}^{n} \binom{2n+1}{k} = 2^{2n} \]

### Step 4: Calculate the Required Sum
From the problem statement:
\[ \sum_{k=1}^{n} \binom{2n+1}{k} = 4095 \]
This can be rewritten as:
\[ \sum_{k=0}^{n} \binom{2n+1}{k} - \binom{2n+1}{0} = 4095 \]
Substituting \( \binom{2n+1}{0} = 1 \):
\[ 2^{2n} - 1 = 4095 \]
Thus:
\[ 2^{2n} = 4096 \]

### Step 5: Solve for \( n \)
Since \( 4096 = 2^{12} \):
\[ 2n = 12 \implies n = 6 \]

### Conclusion for Statement 1
The statement claims \( n = 7 \), which is incorrect. Therefore, Statement 1 is **false**.

### Step 6: Analyze Statement 2
The second statement is:
\[ \binom{n}{r} = \binom{n}{n-r} \]
This is a well-known identity in combinatorics, which states that choosing \( r \) items from \( n \) is the same as choosing \( n-r \) items from \( n \).

### Conclusion for Statement 2
Since this identity is true, Statement 2 is **true**.

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