<p>The number of ways \(= {}^{21}C_0 + {}^{21}C_1 + {}^{21}C_2 + \cdots + {}^{21}C_{10}\)</p><p>\(= \dfrac{2^{22}}{2}\)</p><p>What is this equal to?</p>
Step-by-Step Solution
Key Concept: Use the symmetry property of binomial coefficients: ⁿCᵣ = ⁿCₙ₋ᵣ, and the fact that the sum of all binomial coefficients equals 2ⁿ. Since we're summing exactly half the coefficients (0 to 10 out of 0 to 21), this equals 2²¹/2 = 2²⁰.
<p><strong>Step 1:</strong> Recall that the sum of all binomial coefficients is: ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ = 2ⁿ</p><p>For n = 21: ²¹C₀ + ²¹C₁ + ... + ²¹C₂₁ = 2²¹</p><p><strong>Step 2:</strong> Use the symmetry property ²¹Cᵣ = ²¹C₂₁₋ᵣ. This means:</p><p>²¹C₀ = ²¹C₂₁, ²¹C₁ = ²¹C₂₀, ..., ²¹C₁₀ = ²¹C₁₁</p><p><strong>Step 3:</strong> The sum from r = 0 to 10 contains exactly half of all the pairs (since there are 22 terms total, indices 0 to 21). Therefore:</p><p>²¹C₀ + ²¹C₁ + ... + ²¹C₁₀ = (2²¹)/2 = 2²⁰</p><p><strong>Step 4:</strong> Simplify the given expression 2²²/2:</p><p>2²²/2 = 2²² × 2⁻¹ = 2²¹</p><p>∴ Answer: <strong>2²¹</strong> or equivalently <strong>2,097,152</strong></p>
Correct Answer: C