Example 12 - Find value of n nP4 / n-1P4 = 5/3 and  nP5 = 42 nP3

Example 12 - Chapter 7 Class 11 Permutations and Combinations - Part 2
Example 12 - Chapter 7 Class 11 Permutations and Combinations - Part 3

  1. Chapter 7 Class 11 Permutations and Combinations (Term 2)
  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 = 𝑛!/(𝑛 βˆ’ 5)! = (𝑛(𝑛 βˆ’ 1)(𝑛 βˆ’ 2)(𝑛 βˆ’ 3)(𝑛 βˆ’ 4)(𝑛 βˆ’ 5)!)/(𝑛 βˆ’ 5)! = n(n – 1)(n – 2)(n – 3)(n – 4) Calculating 42nP3 42nP3 = 42𝑛!/(𝑛 βˆ’ 3)! = 42𝑛(𝑛 βˆ’1)(𝑛 βˆ’ 2)(𝑛 βˆ’ 3)!/(𝑛 βˆ’ 3)! = 42n(n – 1)(n – 2) Now, nP5 = 42 nP3 n(n – 1)(n – 2)(n – 3)(n – 4) = 42n(n – 1)(n – 2) (𝑛(𝑛 βˆ’ 1)(𝑛 βˆ’ 2)(𝑛 βˆ’ 3)(𝑛 βˆ’ 4) )/(𝑛(𝑛 βˆ’ 1)(𝑛 βˆ’ 2) ) = 42 (n – 3)(n – 4) = 42 n(n – 4) – 3(n – 4) = 42 n2 – 4n – 3n + 12 = 42 n2 – 7n + 12 = 42 n2 – 7n + 12 – 42 = 0 n2 – 10n + 3n – 30 = 0 n(n – 10) + 3(n – 10) = 0 (n – 10) (n + 3) = 0 So, n = 10, and n = – 3 n(n – 10) + 3(n – 10) = 0 (n – 10) (n + 3) = 0 So, n = 10, and n = –3 But, It is given in question n > 4 So n = –3 not possible Therefore, n = 10 only

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.