<p>Number of ways = <br>\({}^{21}C_0 + {}^{21}C_1 + {}^{21}C_2 + \cdots + {}^{21}C_{10}\)</p><p>Find the value of this sum.</p>
Step-by-Step Solution
Key Concept: Use the symmetry property of binomial coefficients (ⁿCᵣ = ⁿCₙ₋ᵣ) combined with the fact that the sum of all binomial coefficients equals 2ⁿ. The sum from C₀ to C₁₀ equals exactly half of the total sum when n is odd.
<p><strong>Step 1:</strong> Recall that for any positive integer n: $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n$</p><p><strong>Step 2:</strong> For n = 21 (odd): $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$</p><p><strong>Step 3:</strong> Use symmetry property $\binom{21}{r} = \binom{21}{21-r}$. This means:</p><ul><li>$\binom{21}{0} = \binom{21}{21}$</li><li>$\binom{21}{1} = \binom{21}{20}$</li><li>$\binom{21}{k} = \binom{21}{21-k}$ for all k</li></ul><p><strong>Step 4:</strong> The sum splits exactly in half:</p><p>$\binom{21}{0} + \binom{21}{1} + \cdots + \binom{21}{10} = \binom{21}{11} + \binom{21}{12} + \cdots + \binom{21}{21}$</p><p><strong>Step 5:</strong> Therefore: $2\left(\binom{21}{0} + \binom{21}{1} + \cdots + \binom{21}{10}\right) = 2^{21}$</p><p><strong>Step 6:</strong> $\binom{21}{0} + \binom{21}{1} + \cdots + \binom{21}{10} = \frac{2^{21}}{2} = 2^{20}$</p><p>∴ Answer: $\boxed{2^{20}}$</p>
Correct Answer: A