Permutations & Combinations
Committee selection
Grade 11
Question:
<p>There are two possible combinations (1 Boy + 2 Girls) or (2 Boys + 1 Girl). The number of ways \({}^5C_1 \cdot {}^nC_2 + {}^5C_2 \cdot {}^nC_1 = 1750\).</p><p>That is, \(\left(5 \times \dfrac{n(n-1)}{2 \times 1}\right) + \left(\dfrac{5 \times 4}{2} \times n\right) = 1750\)</p><p>Find the value of \(n\).</p>
<p>20</p>
<p>22</p>
<p>24</p>
<p>25</p>
Step-by-Step Solution
Key Concept: Recognize that two mutually exclusive selection cases must be added, then simplify the resulting equation by factoring out common terms to create a solvable quadratic.
<p><strong>Step 1:</strong> Expand both terms:</p><p>Left side: $5 \times \frac{n(n-1)}{2} + 10n = 1750$</p><p>Simplify: $\frac{5n(n-1)}{2} + 10n = 1750$</p><p><strong>Step 2:</strong> Multiply entire equation by 2 to eliminate fraction:</p><p>$5n(n-1) + 20n = 3500$</p><p>$5n^2 - 5n + 20n = 3500$</p><p>$5n^2 + 15n = 3500$</p><p><strong>Step 3:</strong> Factor out 5:</p><p>$5(n^2 + 3n) = 3500$</p><p>$n^2 + 3n = 700$</p><p>$n^2 + 3n - 700 = 0$</p><p><strong>Step 4:</strong> Apply quadratic formula or factorization:</p><p>$(n + 28)(n - 25) = 0$</p><p>$n = -28$ or $n = 25$</p><p><strong>Step 5:</strong> Since n represents the number of girls, $n > 0$</p><p>∴ Answer: $n = 25$</p>
Correct Answer: D