Find the value of `^10C_2 + ^10C_3 + ... + ^10C_10`

Options

A. 100
B. `10^2 - 10`
C. `2^10 - 11`
D. `2^10 - 2`

Answer

To find the value of \( \binom{10}{2} + \binom{10}{3} + \ldots + \binom{10}{10} \), we can use the properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understand the Binomial Theorem**:
The Binomial Theorem states that:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
For our case, we can set \( x = 1 \) and \( y = 1 \):
\[ (1 + 1)^{10} = \sum_{k=0}^{10} \binom{10}{k} = 2^{10} \]

2. **Calculate \( 2^{10} \)**:
\[ 2^{10} = 1024 \]
This means:
\[ \sum_{k=0}^{10} \binom{10}{k} = 1024 \]

3. **Separate the Terms**:
We want to find \( \binom{10}{2} + \binom{10}{3} + \ldots + \binom{10}{10} \). We can express this as:
\[ \sum_{k=2}^{10} \binom{10}{k} = \sum_{k=0}^{10} \binom{10}{k} - \binom{10}{0} - \binom{10}{1} \]

4. **Calculate \( \binom{10}{0} \) and \( \binom{10}{1} \)**:
\[ \binom{10}{0} = 1 \quad \text{and} \quad \binom{10}{1} = 10 \]

5. **Combine the Results**:
Now we substitute back:
\[ \sum_{k=2}^{10} \binom{10}{k} = 1024 - 1 - 10 = 1024 - 11 = 1013 \]

### Final Answer:
Thus, the value of \( \binom{10}{2} + \binom{10}{3} + \ldots + \binom{10}{10} \) is \( 1013 \).

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