Find `""^11C_0 + ""^11C_1 + ""^11C_2 + ……+ ""^11C_5 = ` ?

Options

A. `2^7`
B. `2^8`
C. `2^9`
D. `2^10`

Answer

To solve the problem of finding the sum \( \binom{11}{0} + \binom{11}{1} + \binom{11}{2} + \ldots + \binom{11}{5} \), we can follow these steps:

### Step 1: Understand the Binomial Coefficients
The binomial coefficient \( \binom{n}{r} \) is defined as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
In this case, \( n = 11 \) and \( r \) varies from 0 to 5.

### Step 2: Calculate Each Binomial Coefficient
We will calculate each of the required binomial coefficients:

1. **Calculate \( \binom{11}{0} \)**:
\[ \binom{11}{0} = \frac{11!}{0! \cdot 11!} = 1 \]

2. **Calculate \( \binom{11}{1} \)**:
\[ \binom{11}{1} = \frac{11!}{1! \cdot 10!} = \frac{11}{1} = 11 \]

3. **Calculate \( \binom{11}{2} \)**:
\[ \binom{11}{2} = \frac{11!}{2! \cdot 9!} = \frac{11 \times 10}{2 \times 1} = 55 \]

4. **Calculate \( \binom{11}{3} \)**:
\[ \binom{11}{3} = \frac{11!}{3! \cdot 8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \]

5. **Calculate \( \binom{11}{4} \)**:
\[ \binom{11}{4} = \frac{11!}{4! \cdot 7!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \]

6. **Calculate \( \binom{11}{5} \)**:
\[ \binom{11}{5} = \frac{11!}{5! \cdot 6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \]

### Step 3: Sum the Binomial Coefficients
Now we add all the calculated values:
\[ \binom{11}{0} + \binom{11}{1} + \binom{11}{2} + \binom{11}{3} + \binom{11}{4} + \binom{11}{5} = 1 + 11 + 55 + 165 + 330 + 462 \]

Calculating the sum:
\[ 1 + 11 = 12 \]
\[ 12 + 55 = 67 \]
\[ 67 + 165 = 232 \]
\[ 232 + 330 = 562 \]
\[ 562 + 462 = 1024 \]

### Final Result
Thus, the final result is:
\[ \binom{11}{0} + \binom{11}{1} + \binom{11}{2} + \binom{11}{3} + \binom{11}{4} + \binom{11}{5} = 1024 \]