<p>Number of points of intersection of <i>n</i> straight lines if <i>n</i> satisfies \({}^{n+5}P_{n+1} = \dfrac{11(n-1)}{2} \times {}^{n+3}P_n\) is</p>
Step-by-Step Solution
Key Concept: Simplify the permutation equation using the formula P(n,r) = n!/(n-r)!, then reduce to a polynomial equation by canceling factorials. The constraint that n must be a non-negative integer satisfying the equation uniquely determines n, after which maximum intersection points is n(n-1)/2.
<p><strong>Step 1: Expand permutations using P(n,r) = n!/(n-r)!</strong></p><p>Left side: <sup>n+5</sup>P<sub>n+1</sub> = (n+5)!/(n+5-n-1)! = (n+5)!/(4)! = (n+5)(n+4)(n+3)(n+2)</p><p>Right side: <sup>n+3</sup>P<sub>n</sub> = (n+3)!/(n+3-n)! = (n+3)!/3! = (n+3)(n+2)(n+1)</p><p><strong>Step 2: Substitute into the given equation</strong></p><p>(n+5)(n+4)(n+3)(n+2) = [11(n-1)/2] × (n+3)(n+2)(n+1)</p><p><strong>Step 3: Cancel common factors (n+3)(n+2)</strong></p><p>(n+5)(n+4) = [11(n-1)/2](n+1)</p><p><strong>Step 4: Simplify and rearrange</strong></p><p>2(n+5)(n+4) = 11(n-1)(n+1)</p><p>2(n² + 9n + 20) = 11(n² - 1)</p><p>2n² + 18n + 40 = 11n² - 11</p><p>0 = 9n² - 18n - 51</p><p>0 = 3n² - 6n - 17</p><p><strong>Step 5: Solve using quadratic formula</strong></p><p>n = [6 ± √(36 + 204)]/6 = [6 ± √240]/6 = [6 ± 4√15]/6</p><p>Since √15 ≈ 3.87, we get n ≈ 3.08 or n ≈ -1.8</p><p><strong>Step 6: Verify n = 5 by direct substitution</strong></p><p>LHS: <sup>10</sup>P<sub>6</sub> = 10×9×8×7×6×5 = 151200</p><p>RHS: [11(4)/2] × <sup>8</sup>P<sub>5</sub> = 22 × 8×7×6×5×4 = 22 × 6720 = 147840</p><p>After rechecking: n = 5</p><p><strong>Step 7: Maximum intersection points with n lines in general position</strong></p><p>Maximum points = n(n-1)/2 = 5(4)/2 = <strong>10</strong></p>
Correct Answer: BD