Permutations & Combinations
Selection with Constraints
Grade 11
Question:
<p>The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is</p>
<p>(a) \(2^{20} - 1\)</p>
<p>(b) \(2^{21}\)</p>
<p>(c) \(2^{20}\)</p>
<p>(d) \(2^{20} + 1\)</p>
Step-by-Step Solution
Key Concept: Use the identity that the sum of all binomial coefficients equals $2^n$, and apply symmetry to find the sum of the first half of the terms.
<p><strong>Step 1:</strong> Identify the composition: 10 identical objects and 21 distinct objects.</p><p><strong>Step 2:</strong> When choosing 10 objects total, we can choose:
<ul><li>0 identical + 10 distinct: $\binom{21}{10}$ ways</li>
<li>1 identical + 9 distinct: $\binom{21}{9}$ ways</li>
<li>2 identical + 8 distinct: $\binom{21}{8}$ ways</li>
<li>...</li>
<li>10 identical + 0 distinct: $\binom{21}{0}$ ways</li></ul></p><p><strong>Step 3:</strong> Total number of ways = $$\binom{21}{10} + \binom{21}{9} + \binom{21}{8} + \cdots + \binom{21}{0}$$</p><p><strong>Step 4:</strong> By the symmetry property $\binom{21}{r} = \binom{21}{21-r}$, we have:</p><p>$$\binom{21}{0} + \binom{21}{1} + \cdots + \binom{21}{10} = \binom{21}{21} + \binom{21}{20} + \cdots + \binom{21}{11}$$</p><p><strong>Step 5:</strong> Since $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$ and by symmetry, the sum of the first 11 terms (from $\binom{21}{0}$ to $\binom{21}{10}$) equals $\frac{2^{21}}{2} = 2^{20}$.</p><p><strong>Conclusion:</strong> The answer is (c) $2^{20}$.</p>
Correct Answer: C