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 find the value of 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: Rewrite the expression

We can group the terms in the expression based on the symmetry of binomial coefficients:

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

Notice that:

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

Thus, we can rewrite the expression as:

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

### Step 2: Use the Binomial Theorem

The Binomial Theorem states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

For \( a = 1 \) and \( b = 1 \), we have:

\[ (1 + 1)^{20} = \sum_{k=0}^{20} \binom{20}{k} = 2^{20} \]

### Step 3: Calculate the total sum

The total sum of all the coefficients from \( k = 0 \) to \( k = 20 \) is \( 2^{20} \).

### Step 4: Split the sum

The sum of the coefficients can be split into two halves:

\[ \sum_{k=0}^{20} \binom{20}{k} = \sum_{k=0}^{10} \binom{20}{k} + \sum_{k=11}^{20} \binom{20}{k} \]

By symmetry, we know:

\[ \sum_{k=0}^{10} \binom{20}{k} = \sum_{k=11}^{20} \binom{20}{k} = \frac{2^{20}}{2} = 2^{19} \]

### Step 5: Calculate the required sum

Now, we need to find the sum of the coefficients from \( k = 0 \) to \( k = 4 \) and from \( k = 12 \) to \( k = 15 \):

\[ \sum_{k=0}^{4} \binom{20}{k} + \sum_{k=12}^{15} \binom{20}{k} \]

We can express the sum from \( k = 5 \) to \( k = 11 \) as:

\[ \sum_{k=5}^{11} \binom{20}{k} = \sum_{k=0}^{10} \binom{20}{k} - \sum_{k=0}^{4} \binom{20}{k} \]

### Step 6: Final calculation

Thus, we can write:

\[ \text{Required Sum} = 2^{19} - \left( \binom{20}{5} + \binom{20}{6} + \binom{20}{7} + \binom{20}{8} + \binom{20}{9} + \binom{20}{10} + \binom{20}{11} \right) \]

### Conclusion

The final answer can be computed using the values of the binomial coefficients, but we can conclude that the value of the original expression is:

\[ \text{Value} = 2^{19} - \left( \binom{20}{9} + \binom{20}{10} + \binom{20}{11} \right) \]