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  1. Chapter 7 Class 11 Permutations and Combinations
  2. Serial order wise

Transcript

Example 12 Find the value of n such that nP5 = 42 nP3, n > 4 Given nP5 = 42 nP3 Calculating nP5 nP5 = ๐‘›!/(๐‘› โˆ’ 5)! = (๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2)(๐‘› โˆ’ 3)(๐‘› โˆ’ 4)(๐‘› โˆ’ 5)!)/(๐‘› โˆ’ 5)! = n(n โ€“ 1)(n โ€“ 2)(n โ€“ 3)(n โ€“ 4) Calculating 42nP3 42nP3 = 42๐‘›!/(๐‘› โˆ’ 3)! = 42๐‘›(๐‘› โˆ’1)(๐‘› โˆ’ 2)(๐‘› โˆ’ 3)!/(๐‘› โˆ’ 3)! = 42n(n โ€“ 1)(n โ€“ 2) Now, nP5 = 42 nP3 n(n โ€“ 1)(n โ€“ 2)(n โ€“ 3)(n โ€“ 4) = 42n(n โ€“ 1)(n โ€“ 2) (๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2)(๐‘› โˆ’ 3)(๐‘› โˆ’ 4) )/(๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2) ) = 42 (n โ€“ 3)(n โ€“ 4) = 42 n(n โ€“ 4) โ€“ 3(n โ€“ 4) = 42 n2 โ€“ 4n โ€“ 3n + 12 = 42 n2 โ€“ 7n + 12 = 42 n2 โ€“ 7n + 12 โ€“ 42 = 0 n2 โ€“ 10n + 3n โ€“ 30 = 0 n(n โ€“ 10) + 3(n โ€“ 10) = 0 (n โ€“ 10) (n + 3) = 0 So, n = 10, and n = โ€“ 3 n(n โ€“ 10) + 3(n โ€“ 10) = 0 (n โ€“ 10) (n + 3) = 0 So, n = 10, and n = โ€“3 But, It is given in question n > 4 So n = โ€“3 not possible Therefore, n = 10 only Example 12 Find the value of n such that (ii) "nP4" /"nโˆ’1P4" = 5/3 , n > 4 Lets first calculate nP4 and n โ€“ 1P4 nP4 = ๐‘›!/(๐‘› โˆ’ 4)! = (๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2)(๐‘› โˆ’ 3)(๐‘› โˆ’ 4)!)/(๐‘› โˆ’ 4)! = n(n โ€“ 1)(n โ€“ 2)(n โ€“ 3) n โ€“ 1P4 = ((๐‘› โˆ’ 1)!)/(๐‘› โˆ’ 1 โˆ’ 4)! = ((๐‘› โˆ’ 1)!)/(๐‘› โˆ’ 5)! = ((๐‘› โˆ’ 1)(๐‘› โˆ’ 2)(๐‘› โˆ’ 3)(๐‘› โˆ’ 4)(๐‘› โˆ’ 5)!)/(๐‘› โˆ’ 5)! = (n โ€“ 1) (n โ€“ 2) (n โ€“ 3) (n โ€“ 4) nPr = ((๐‘›)!)/(๐‘› โˆ’ ๐‘Ÿ)! Now "nP4" /"nโˆ’1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n โ€“ 1)(n โ€“ 2)(n โ€“ 3) = 5(n โ€“ 1) (n โ€“ 2) (n โ€“ 3) (n โ€“ 4) (3๐‘›(๐‘› โˆ’1)(๐‘› โˆ’2)(๐‘› โˆ’ 3))/((๐‘› โˆ’1)(๐‘› โˆ’2)(๐‘› โˆ’ 3)) = 5(n โ€“ 4) 3n = 5(n โ€“ 4) 3n = 5n โ€“ 20 20 = 5n โ€“ 3n 20 = 2n 20/2 = n 10 = n Hence, n = 10 n(n โ€“ 10) + 3(n โ€“ 10) = 0 (n โ€“ 10) (n + 3) = 0 So, n = 10, and n = โ€“3 But, It is given in question n > 4 So n = โ€“3 not possible Therefore, n = 10 only Example 12 Find the value of n such that (ii) "nP4" /"nโˆ’1P4" = 5/3 , n > 4 Lets first calculate nP4 and n โ€“ 1P4 nP4 = ๐‘›!/(๐‘› โˆ’ 4)! = (๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2)(๐‘› โˆ’ 3)(๐‘› โˆ’ 4)!)/(๐‘› โˆ’ 4)! = n(n โ€“ 1)(n โ€“ 2)(n โ€“ 3) n โ€“ 1P4 = ((๐‘› โˆ’ 1)!)/(๐‘› โˆ’ 1 โˆ’ 4)! = ((๐‘› โˆ’ 1)!)/(๐‘› โˆ’ 5)! = ((๐‘› โˆ’ 1)(๐‘› โˆ’ 2)(๐‘› โˆ’ 3)(๐‘› โˆ’ 4)(๐‘› โˆ’ 5)!)/(๐‘› โˆ’ 5)! = (n โ€“ 1) (n โ€“ 2) (n โ€“ 3) (n โ€“ 4) nPr = ((๐‘›)!)/(๐‘› โˆ’ ๐‘Ÿ)! Now "nP4" /"nโˆ’1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n โ€“ 1)(n โ€“ 2)(n โ€“ 3) = 5(n โ€“ 1) (n โ€“ 2) (n โ€“ 3) (n โ€“ 4) (3๐‘›(๐‘› โˆ’1)(๐‘› โˆ’2)(๐‘› โˆ’ 3))/((๐‘› โˆ’1)(๐‘› โˆ’2)(๐‘› โˆ’ 3)) = 5(n โ€“ 4) 3n = 5(n โ€“ 4) 3n = 5n โ€“ 20 20 = 5n โ€“ 3n 20 = 2n 20/2 = n 10 = n Hence, n = 10

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.