Question

The value of ${}^{11}C_{0}+{}^{11}C_{1}+{}^{11}C_{2}+{}^{11}C_{3}+{}^{11}C_{4}+{}^{11}C_{5}$ is

• A. $2^{6}$
• B. $2^{8}$
• C. $2^{10}$
• D. $2^{11}$
• E. $2^{9}$

Hint:

Binomial Theorem Tip: For any odd $n$, the sum of the first half of the binomial coefficients (from $r=0$ to $r=\frac{n-1}{2}$) is always $2^{n-1}$.

Answer 1

The Correct Option is C

Concept:
The sum of all binomial coefficients for a given power $n$ is $\sum_{r=0}^{n} {}^{n}C_r = 2^n$. Additionally, binomial coefficients are symmetric, meaning ${}^{n}C_r = {}^{n}C_{n-r}$. When $n$ is odd, the sum of the first half of the coefficients is exactly equal to the sum of the second half.

Step 1: Write the full sum of coefficients.
For $n = 11$, the sum of all binomial coefficients from $r=0$ to $r=11$ is: $${}^{11}C_0 + {}^{11}C_1 + \dots + {}^{11}C_{10} + {}^{11}C_{11} = 2^{11}$$

Step 2: Apply the symmetry property.
Because ${}^{11}C_r = {}^{11}C_{11-r}$, we can pair the terms: ${}^{11}C_0 = {}^{11}C_{11}$ ${}^{11}C_1 = {}^{11}C_{10}$ ... ${}^{11}C_5 = {}^{11}C_6$

Step 3: Split the full sum into two equal halves.
The entire sequence can be divided into a left half and a right half: Left half: $S = {}^{11}C_0 + {}^{11}C_1 + {}^{11}C_2 + {}^{11}C_3 + {}^{11}C_4 + {}^{11}C_5$ Right half: $S = {}^{11}C_6 + {}^{11}C_7 + {}^{11}C_8 + {}^{11}C_9 + {}^{11}C_{10} + {}^{11}C_{11}$

Step 4: Set up the algebraic equation.
Since the left half equals the right half, their sum is $2S$: $$S + S = 2^{11}$$ $$2S = 2^{11}$$

Step 5: Solve for the requested sum S.
Divide both sides by 2 to isolate S: $$S = \frac{2^{11}}{2}$$ $$S = 2^{11-1} = 2^{10}$$ Hence the correct answer is (C) $2^{10$}.