The sum `""^(20)C_(0)+""^(20)C_(1)+""^(20)C_(2)+……+""^(20)C_(10)` is equal to :

Options

A. `2^(20)+(20!)/(10!)^(2)`
B. `2^(19)-1/2*(20!)/(10!)^(2)`
C. `2^(19)+""^(20)C_(10)`
D. None of these

Answer

To solve the problem of finding the sum \( \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \ldots + \binom{20}{10} \), we can use properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Sum**:
We need to calculate the sum of the first half of the binomial coefficients for \( n = 20 \). This can be expressed as:
\[ S = \sum_{k=0}^{10} \binom{20}{k} \]

2. **Using the Symmetry of Binomial Coefficients**:
The binomial coefficients have a symmetry property:
\[ \binom{n}{k} = \binom{n}{n-k} \]
For \( n = 20 \), this means:
\[ \binom{20}{0} = \binom{20}{20}, \quad \binom{20}{1} = \binom{20}{19}, \quad \ldots, \quad \binom{20}{10} = \binom{20}{10} \]
Therefore, we can express the sum of all coefficients from \( k = 0 \) to \( k = 20 \) as:
\[ \sum_{k=0}^{20} \binom{20}{k} = 2^{20} \]

3. **Dividing the Total Sum**:
The total sum \( \sum_{k=0}^{20} \binom{20}{k} = 2^{20} \) includes both halves of the coefficients:
\[ \sum_{k=0}^{10} \binom{20}{k} + \sum_{k=11}^{20} \binom{20}{k} = 2^{20} \]
By the symmetry property:
\[ \sum_{k=0}^{10} \binom{20}{k} = \sum_{k=11}^{20} \binom{20}{k} \]
Let \( S = \sum_{k=0}^{10} \binom{20}{k} \). Then:
\[ S + S = 2S = 2^{20} \]
Thus:
\[ S = \frac{2^{20}}{2} = 2^{19} \]

4. **Final Result**:
Therefore, the sum \( \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \ldots + \binom{20}{10} \) is equal to:
\[ S = 2^{19} \]

### Conclusion:
The answer to the problem is:
\[ \boxed{2^{19}} \]