`""^21C_0 + ""^21C_1 + ""^21C_2 + ……..+ ""^21C_10 = `

Options

A. `2^10`
B. `2^20`
C. `2^21`
D. `2^19`

Answer

To solve the problem \( \binom{21}{0} + \binom{21}{1} + \binom{21}{2} + \ldots + \binom{21}{10} \), we will use the Binomial Theorem and properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The Binomial Theorem states that:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
For our case, we will set \( n = 21 \).

2. **Writing the Expansion**:
We can write the expansion for \( (1 + x)^{21} \):
\[ (1 + x)^{21} = \binom{21}{0} x^0 + \binom{21}{1} x^1 + \binom{21}{2} x^2 + \ldots + \binom{21}{21} x^{21} \]

3. **Substituting \( x = 1 \)**:
Now, we substitute \( x = 1 \) into the expansion:
\[ (1 + 1)^{21} = 2^{21} = \binom{21}{0} + \binom{21}{1} + \binom{21}{2} + \ldots + \binom{21}{21} \]
This gives us the sum of all the binomial coefficients from \( \binom{21}{0} \) to \( \binom{21}{21} \).

4. **Grouping the Coefficients**:
Notice that the binomial coefficients have a symmetry:
\[ \binom{21}{k} = \binom{21}{21-k} \]
This means:
\[ \binom{21}{0} = \binom{21}{21}, \quad \binom{21}{1} = \binom{21}{20}, \quad \ldots, \quad \binom{21}{10} = \binom{21}{11} \]

5. **Finding the Required Sum**:
Since the coefficients from \( \binom{21}{0} \) to \( \binom{21}{10} \) are equal to the coefficients from \( \binom{21}{11} \) to \( \binom{21}{21} \), we can say:
\[ \binom{21}{0} + \binom{21}{1} + \ldots + \binom{21}{10} = \binom{21}{11} + \binom{21}{12} + \ldots + \binom{21}{21} \]
Therefore, we can write:
\[ \binom{21}{0} + \binom{21}{1} + \ldots + \binom{21}{10} + \binom{21}{11} + \ldots + \binom{21}{21} = 2^{21} \]
This implies:
\[ 2 \left( \binom{21}{0} + \binom{21}{1} + \ldots + \binom{21}{10} \right) = 2^{21} \]

6. **Dividing by 2**:
To find the sum we want, we divide both sides by 2:
\[ \binom{21}{0} + \binom{21}{1} + \ldots + \binom{21}{10} = \frac{2^{21}}{2} = 2^{20} \]

### Final Answer:
\[ \binom{21}{0} + \binom{21}{1} + \ldots + \binom{21}{10} = 2^{20} \]