Permutations & Combinations
Combinations and Binomial coefficients
Grade 11
Question:
<p>We have <br/> \(\left({}^{21}C_1 + {}^{21}C_2 + \cdots + {}^{21}C_{10}\right)\left({}^{10}C_1 + {}^{10}C_2 + \cdots + {}^{10}C_{10}\right)\)</p><p>Which of the following is the simplified value of the above expression?</p>
<p>\(2^{20} - 2^{10}\)</p>
<p>\(2^{21} - 2^{10}\)</p>
<p>\(2^{20} - 2^{11}\)</p>
<p>\(2^{21} - 2^{11}\)</p>
Step-by-Step Solution
Key Concept: Use the binomial theorem identity: sum of binomial coefficients equals 2^n, and recognize that the first sum is half of (2^21 - 2) due to symmetry of Pascal's triangle, while the second sum equals 2^10 - 1.
<p><strong>Step 1:</strong> Simplify the first sum using the binomial theorem.</p><p>We know: $2^{21} = \binom{21}{0} + \binom{21}{1} + \binom{21}{2} + \cdots + \binom{21}{21}$</p><p>By symmetry: $\binom{21}{1} + \binom{21}{2} + \cdots + \binom{21}{10} = \frac{2^{21} - 2}{2} = 2^{20} - 1$</p><p>(We subtract 2 for $\binom{21}{0}$ and $\binom{21}{21}$, then divide by 2 due to symmetry)</p><p><strong>Step 2:</strong> Simplify the second sum.</p><p>We know: $2^{10} = \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \cdots + \binom{10}{10}$</p><p>Therefore: $\binom{10}{1} + \binom{10}{2} + \cdots + \binom{10}{10} = 2^{10} - \binom{10}{0} = 2^{10} - 1$</p><p><strong>Step 3:</strong> Multiply the simplified expressions.</p><p>$(2^{20} - 1)(2^{10} - 1) = 2^{30} - 2^{20} - 2^{10} + 1$</p><p>∴ Answer: A</p>
Correct Answer: A