With usual notations,
`C_(0)C_(r )+C_(1)C_(r+1) +C_(2)C_(r+2) +…+C_(n-r)C_(n)=`

Options

A. `((2n)!)/((n!)^(2))`
B. `((2n)!)/((n-r)! (n+r)!)`
C. `((2n)!)/([(n-r)!]^(2))`
D. none

Answer

To solve the given expression \( C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n \), we will use the Binomial Theorem and properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression involves binomial coefficients \( C_k \), which is defined as \( C_k = \binom{n}{k} \). The goal is to simplify the sum \( S = C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n \).

2. **Using the Binomial Theorem**:
Recall the Binomial Theorem states:
\[ (x + y)^n = \sum_{k=0}^{n} C_k x^k y^{n-k} \]
We can apply this theorem to two different expansions:
- \( (1 + x)^n = \sum_{k=0}^{n} C_k x^k \)
- \( (1 + y)^n = \sum_{k=0}^{n} C_k y^k \)

3. **Multiplying the Two Expansions**:
We can multiply the two expansions:
\[ (1 + x)^n (1 + y)^n = \sum_{k=0}^{n} C_k x^k \sum_{j=0}^{n} C_j y^j \]
This results in:
\[ = \sum_{m=0}^{2n} \left( \sum_{k=0}^{m} C_k C_{m-k} \right) x^k y^{m-k} \]

4. **Identifying the Coefficient**:
To find the coefficient of \( x^{r} y^{n-r} \) in the expanded product, we set \( x = 1 \) and \( y = 1 \):
\[ (1 + 1)^n (1 + 1)^n = 2^n \cdot 2^n = 4^n \]
The coefficient of \( x^{n-r} \) in this expansion corresponds to our original expression.

5. **Finding the Coefficient**:
The coefficient of \( x^{n-r} \) in the expansion of \( (1 + x)^{2n} \) is given by \( C_{n-r} \) from the expansion:
\[ (1 + x)^{2n} = \sum_{k=0}^{2n} C_k x^k \]
Thus, the coefficient of \( x^{n-r} \) is \( C_{n-r} \).

6. **Final Result**:
Therefore, we conclude that:
\[ S = C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = C_{n-r} \]

### Final Answer:
\[ C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = C_{n} \]