Permutations & Combinations
Properties of Combinations
Grade None
Question:
<p>If \({}^{15}C_{3r} : {}^{15}C_{r+1} = 11:3\), find the value of <em>r</em>.</p>
Step-by-Step Solution
Key Concept: Use the combination formula ratio property: $\frac{{}^nC_a}{{}^nC_b} = \frac{(n-a)!(b)!}{(a)!(n-b)!}$ to convert the ratio into a solvable equation. Recognize that when the upper index is same, the ratio simplifies to a factorial expression in terms of the lower indices.
<p><strong>Step 1:</strong> Write the given ratio using combination formula.</p><p>$$\frac{{}^{15}C_{3r}}{{}^{15}C_{r+1}} = \frac{11}{3}$$</p><p><strong>Step 2:</strong> Apply the combination ratio formula.</p><p>$$\frac{{}^{15}C_{3r}}{{}^{15}C_{r+1}} = \frac{\frac{15!}{(3r)!(15-3r)!}}{\frac{15!}{(r+1)!(15-r-1)!}} = \frac{(r+1)!(14-r)!}{(3r)!(15-3r)!}$$</p><p><strong>Step 3:</strong> Simplify using factorial properties.</p><p>$$\frac{(r+1)·r·(r-1)·...·1 · (14-r)·(13-r)·...·1}{(3r)·(3r-1)·...·1 · (15-3r)·(14-3r)·...·1} = \frac{11}{3}$$</p><p><strong>Step 4:</strong> Test small integer values of r systematically.</p><p>For <strong>r = 3</strong>:</p><p>$$\frac{{}^{15}C_9}{{}^{15}C_4} = \frac{\frac{15!}{9!·6!}}{\frac{15!}{4!·11!}} = \frac{4!·11!}{9!·6!} = \frac{24·39916800}{362880·720} = \frac{11}{3}$$ ✓</p><p>Verify: ${}^{15}C_9 = 5005$ and ${}^{15}C_4 = 1365$; ratio = $\frac{5005}{1365} = \frac{11}{3}$ ✓</p><p>∴ <strong>Answer: r = 3</strong></p>
Correct Answer: 3