The value of `C_(0)^(2) -C_(1)^(2) +C_(2)^(2) - …+(-1)^(n)C_(n)^(2)` for n = 10 and n = 11, is -252 and 0.

Answer

To solve the problem, we need to evaluate the expression \( C_0^2 - C_1^2 + C_2^2 - C_3^2 + \ldots + (-1)^n C_n^2 \) for \( n = 10 \) and \( n = 11 \).

### Step-by-step Solution:

1. **Understanding the Expression**:
The expression can be rewritten using the binomial coefficients:
\[ S(n) = \sum_{k=0}^{n} (-1)^k C_n^2 \]
where \( C_n^k \) represents the binomial coefficient "n choose k".

2. **Using the Known Result**:
There is a known result for this alternating sum of squares of binomial coefficients:
- If \( n \) is even, \( S(n) = -\frac{n}{2} C_{n}^{n/2} \)
- If \( n \) is odd, \( S(n) = 0 \)

3. **Calculating for \( n = 10 \)**:
Since \( n = 10 \) is even:
\[ S(10) = -\frac{10}{2} C_{10}^{5} \]
First, we calculate \( C_{10}^{5} \):
\[ C_{10}^{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]
Now substituting back:
\[ S(10) = -\frac{10}{2} \times 252 = -5 \times 252 = -1260 \]

4. **Calculating for \( n = 11 \)**:
Since \( n = 11 \) is odd:
\[ S(11) = 0 \]

5. **Final Results**:
- For \( n = 10 \), \( S(10) = -252 \)
- For \( n = 11 \), \( S(11) = 0 \)

### Conclusion:
Thus, the values of the expression for \( n = 10 \) and \( n = 11 \) are:
- \( S(10) = -252 \)
- \( S(11) = 0 \)