Examples

Chapter 7 Class 11 Permutations and Combinations
Serial order wise

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Example 12 Find the value of n such that nP5 = 42 nP3, n > 4 Given nP5 = 42 nP3 Calculating nP5 nP5 = π!/(π β 5)! = (π(π β 1)(π β 2)(π β 3)(π β 4)(π β 5)!)/(π β 5)! = n(n β 1)(n β 2)(n β 3)(n β 4) Calculating 42nP3 42nP3 = 42π!/(π β 3)! = 42π(π β1)(π β 2)(π β 3)!/(π β 3)! = 42n(n β 1)(n β 2) Now, nP5 = 42 nP3 n(n β 1)(n β 2)(n β 3)(n β 4) = 42n(n β 1)(n β 2) (π(π β 1)(π β 2)(π β 3)(π β 4) )/(π(π β 1)(π β 2) ) = 42 (n β 3)(n β 4) = 42 n(n β 4) β 3(n β 4) = 42 n2 β 4n β 3n + 12 = 42 n2 β 7n + 12 = 42 n2 β 7n + 12 β 42 = 0 n2 β 10n + 3n β 30 = 0 n(n β 10) + 3(n β 10) = 0 (n β 10) (n + 3) = 0 So, n = 10, and n = β 3 n(n β 10) + 3(n β 10) = 0 (n β 10) (n + 3) = 0 So, n = 10, and n = β3 But, It is given in question n > 4 So n = β3 not possible Therefore, n = 10 only