<p>(b) If \({}^{56}P_{r+6} : {}^{54}P_{r+3} = 30800:1\), find \(r\).</p>
Step-by-Step Solution
Key Concept: Expand permutation formulas using P(n,r) = n!/(n-r)! and simplify the ratio by canceling factorials; the key is recognizing that the ratio reduces to a product of consecutive integers that must equal 30800.
<p><strong>Step 1:</strong> Write the permutation ratio using the formula P(n,r) = n!/(n-r)!</p><p>$$\frac{{}^{56}P_{r+6}}{{}^{54}P_{r+3}} = \frac{\frac{56!}{(50-r)!}}{\frac{54!}{(51-r)!}} = \frac{56!}{(50-r)!} × \frac{(51-r)!}{54!}$$</p><p><strong>Step 2:</strong> Simplify by expanding factorials:</p><p>$$= \frac{56 × 55 × 54!}{(50-r)!} × \frac{(51-r)!}{54!} = \frac{56 × 55 × (51-r)!}{(50-r)!}$$</p><p><strong>Step 3:</strong> Expand (51-r)! in the numerator:</p><p>$$(51-r)! = (51-r)(50-r)(49-r)...(1)$$</p><p>So: $$\frac{56 × 55 × (51-r)(50-r)!}{(50-r)!} = 56 × 55 × (51-r) = 30800$$</p><p><strong>Step 4:</strong> Solve for (51-r):</p><p>$$56 × 55 × (51-r) = 30800$$</p><p>$$3080 × (51-r) = 30800$$</p><p>$$(51-r) = 10$$</p><p>$$r = 41$$</p><p>∴ Answer: <strong>41</strong></p>
Correct Answer: 41