`""^11C_0^2 - ""^11C_1^2 + ""^11C_2^2 - ""^11C_3^2 + ……- "^11C_11^2 = `

Options

A. 0
B. `""^22C_11`
C. `(-1)^11. ""^22C_11`
D. 1

Answer

To solve the problem \( \binom{11}{0}^2 - \binom{11}{1}^2 + \binom{11}{2}^2 - \binom{11}{3}^2 + \ldots - \binom{11}{11}^2 \), we will use the properties of binomial coefficients and the alternating sum.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression can be rewritten as:
\[ S = \sum_{k=0}^{11} (-1)^k \binom{11}{k}^2 \]
This represents an alternating sum of the squares of the binomial coefficients.

**Hint**: Recognize that the expression involves alternating signs and squares of binomial coefficients.

2. **Using a Known Identity**:
There is a known identity for the sum of squares of binomial coefficients:
\[ \sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 = (-1)^{n} \binom{n}{n/2} \quad \text{if } n \text{ is even, and } 0 \text{ if } n \text{ is odd.} \]

**Hint**: Look for identities involving binomial coefficients and alternating sums.

3. **Determine the Parity of \( n \)**:
In our case, \( n = 11 \), which is odd. According to the identity, if \( n \) is odd, the sum evaluates to 0:
\[ S = 0 \]

**Hint**: Check whether \( n \) is odd or even to apply the correct identity.

4. **Final Result**:
Therefore, the value of the given expression is:
\[ \boxed{0} \]

**Hint**: Conclude by stating the final result clearly.