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Example 6 (ii)
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Example 12 (i) Important
Example 12 (ii) You are here
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Examples
Last updated at Aug. 27, 2021 by Teachoo
Example 12 Find the value of n such that (ii) "nP4" /"nβ1P4" = 5/3 , n > 4 Lets first calculate nP4 and n β 1P4 nP4 = π!/(π β 4)! = (π(π β 1)(π β 2)(π β 3)(π β 4)!)/(π β 4)! = n(n β 1)(n β 2)(n β 3) n β 1P4 = ((π β 1)!)/(π β 1 β 4)! = ((π β 1)!)/(π β 5)! = ((π β 1)(π β 2)(π β 3)(π β 4)(π β 5)!)/(π β 5)! = (n β 1) (n β 2) (n β 3) (n β 4) nPr = ((π)!)/(π β π)! Now "nP4" /"nβ1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n β 1)(n β 2)(n β 3) = 5(n β 1) (n β 2) (n β 3) (n β 4) (3π(π β1)(π β2)(π β 3))/((π β1)(π β2)(π β 3)) = 5(n β 4) 3n = 5(n β 4) 3n = 5n β 20 20 = 5n β 3n 20 = 2n 20/2 = n 10 = n Hence, n = 10 n(n β 10) + 3(n β 10) = 0 (n β 10) (n + 3) = 0 So, n = 10, and n = β3 But, It is given in question n > 4 So n = β3 not possible Therefore, n = 10 only Example 12 Find the value of n such that (ii) "nP4" /"nβ1P4" = 5/3 , n > 4 Lets first calculate nP4 and n β 1P4 nP4 = π!/(π β 4)! = (π(π β 1)(π β 2)(π β 3)(π β 4)!)/(π β 4)! = n(n β 1)(n β 2)(n β 3) n β 1P4 = ((π β 1)!)/(π β 1 β 4)! = ((π β 1)!)/(π β 5)! = ((π β 1)(π β 2)(π β 3)(π β 4)(π β 5)!)/(π β 5)! = (n β 1) (n β 2) (n β 3) (n β 4) nPr = ((π)!)/(π β π)! Now "nP4" /"nβ1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n β 1)(n β 2)(n β 3) = 5(n β 1) (n β 2) (n β 3) (n β 4) (3π(π β1)(π β2)(π β 3))/((π β1)(π β2)(π β 3)) = 5(n β 4) 3n = 5(n β 4) 3n = 5n β 20 20 = 5n β 3n 20 = 2n 20/2 = n 10 = n Hence, n = 10