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 \( S = \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \ldots + \binom{20}{10} \), we can use properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Sum**:
The sum we want to evaluate is:
\[ S = \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \ldots + \binom{20}{10} \]

2. **Using the Binomial Theorem**:
According to the binomial theorem, the sum of all binomial coefficients for a given \( n \) is:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
For \( n = 20 \):
\[ \sum_{k=0}^{20} \binom{20}{k} = 2^{20} \]

3. **Symmetry of Binomial Coefficients**:
The binomial coefficients have a symmetry property:
\[ \binom{n}{k} = \binom{n}{n-k} \]
This means:
\[ \binom{20}{0} = \binom{20}{20}, \quad \binom{20}{1} = \binom{20}{19}, \quad \ldots, \quad \binom{20}{10} = \binom{20}{10} \]

4. **Splitting the Sum**:
The total sum can be split into two equal halves:
\[ \sum_{k=0}^{20} \binom{20}{k} = \sum_{k=0}^{10} \binom{20}{k} + \sum_{k=11}^{20} \binom{20}{k} \]
Since \( \sum_{k=11}^{20} \binom{20}{k} = \sum_{k=0}^{9} \binom{20}{k} \) due to symmetry, we have:
\[ \sum_{k=0}^{20} \binom{20}{k} = 2 \sum_{k=0}^{10} \binom{20}{k} \]
Thus:
\[ 2S = 2^{20} \]

5. **Finding S**:
Dividing both sides by 2 gives:
\[ S = \frac{2^{20}}{2} = 2^{19} \]

6. **Final Result**:
Therefore, the sum \( S \) is:
\[ S = 2^{19} \]