The value of `.^(n)C_(0).^(n)C_(n)+.^(n)C_(1).^(n)C_(n-1)+...+.^(n)C_(n).^(n)C_(0)` is

Options

A. `.^(2n)C_(n-1)`
B. `.^(2n)C_(n)`
C. `.^(2n)C_(n+1)`
D. `.^(2n)C_(2)`

Answer

To solve the problem, we need to find the value of the expression:

\[ \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k} \]

### Step-by-step Solution:

1. **Understanding the Binomial Coefficient:**
The binomial coefficient \(\binom{n}{k}\) represents the number of ways to choose \(k\) elements from a set of \(n\) elements. A key property of binomial coefficients is that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
This means that \(\binom{n}{k}\) and \(\binom{n}{n-k}\) are equal.

2. **Rewriting the Expression:**
Using the property mentioned above, we can rewrite the expression as:
\[ \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k} = \sum_{k=0}^{n} \binom{n}{k} \binom{n}{k} \]
This simplifies to:
\[ \sum_{k=0}^{n} \binom{n}{k}^2 \]

3. **Using the Binomial Theorem:**
We know from the binomial theorem that:
\[ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
If we set \(x = 1\), we get:
\[ (1+1)^n = 2^n = \sum_{k=0}^{n} \binom{n}{k} \]

4. **Finding the Sum of Squares of Binomial Coefficients:**
To find \(\sum_{k=0}^{n} \binom{n}{k}^2\), we can use the identity:
\[ \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \]
This identity states that the sum of the squares of the binomial coefficients for a given \(n\) is equal to the binomial coefficient of \(2n\) choose \(n\).

5. **Final Calculation:**
Thus, we have:
\[ \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k} = \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \]

### Conclusion:
The value of the expression is:

\[ \binom{2n}{n} \]