The value of `.^(20)C_(10)+.^(20)C_(1)+.^(20)C_(2)+.^(20)C_(3)+.^(20)C_(4)+.^(20)C_(12)+.^(20)C_(13)+.^(20)C_(14)+.^(20)C_(15)` is

Options

A. `2^(19)-((.^(20)C_(10)+.^(20)C_(9)))/(2)`
B. `2^(19)-((.^(20)C_(10)+2xx.^(20)C_(9)))/(2)`
C. `2^(19)-(.^(20)C_(10))/(2)`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression:

\[ \binom{20}{10} + \binom{20}{1} + \binom{20}{2} + \binom{20}{3} + \binom{20}{4} + \binom{20}{12} + \binom{20}{13} + \binom{20}{14} + \binom{20}{15} \]

### Step 1: Rearranging the Terms
Notice that the binomial coefficient has a symmetry property:

\[ \binom{n}{k} = \binom{n}{n-k} \]

Using this property, we can rewrite some of the terms:

\[ \binom{20}{12} = \binom{20}{8}, \quad \binom{20}{13} = \binom{20}{7}, \quad \binom{20}{14} = \binom{20}{6}, \quad \binom{20}{15} = \binom{20}{5} \]

Thus, we can rewrite the expression as:

\[ \binom{20}{10} + \binom{20}{1} + \binom{20}{2} + \binom{20}{3} + \binom{20}{4} + \binom{20}{8} + \binom{20}{7} + \binom{20}{6} + \binom{20}{5} \]

### Step 2: Grouping the Terms
Now, we can group the terms:

\[ \left( \binom{20}{1} + \binom{20}{19} \right) + \left( \binom{20}{2} + \binom{20}{18} \right) + \left( \binom{20}{3} + \binom{20}{17} \right) + \left( \binom{20}{4} + \binom{20}{16} \right) + \left( \binom{20}{5} + \binom{20}{15} \right) + \binom{20}{10} \]

### Step 3: 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 \( n = 20 \):

\[ \sum_{k=0}^{20} \binom{20}{k} = 2^{20} \]

### Step 4: Excluding Unwanted Terms
We need to exclude the terms \( \binom{20}{9} \), \( \binom{20}{10} \), and \( \binom{20}{11} \) from the total sum:

\[ \sum_{k=0}^{20} \binom{20}{k} - \left( \binom{20}{9} + \binom{20}{10} + \binom{20}{11} \right) \]

### Step 5: Calculating the Final Value
Thus, the value we need is:

\[ 2^{20} - \left( \binom{20}{9} + \binom{20}{10} + \binom{20}{11} \right) \]

### Step 6: Simplifying Further
Since \( \binom{20}{9} = \binom{20}{11} \) and \( \binom{20}{10} \) is the middle term, we can express this as:

\[ 2^{20} - 2\binom{20}{10} \]

### Final Answer
Now, we can compute the value:

\[ \text{Final Value} = 2^{20} - 2 \cdot \binom{20}{10} \]