Check sibling questions

Example 12 - Chapter 7 Class 11 Permutations and Combinations - Part 4

Example 12 - Chapter 7 Class 11 Permutations and Combinations - Part 5


Transcript

Example 12 Find the value of n such that (ii) "nP4" /"nβˆ’1P4" = 5/3 , n > 4 Lets first calculate nP4 and n – 1P4 nP4 = 𝑛!/(𝑛 βˆ’ 4)! = (𝑛(𝑛 βˆ’ 1)(𝑛 βˆ’ 2)(𝑛 βˆ’ 3)(𝑛 βˆ’ 4)!)/(𝑛 βˆ’ 4)! = n(n – 1)(n – 2)(n – 3) n – 1P4 = ((𝑛 βˆ’ 1)!)/(𝑛 βˆ’ 1 βˆ’ 4)! = ((𝑛 βˆ’ 1)!)/(𝑛 βˆ’ 5)! = ((𝑛 βˆ’ 1)(𝑛 βˆ’ 2)(𝑛 βˆ’ 3)(𝑛 βˆ’ 4)(𝑛 βˆ’ 5)!)/(𝑛 βˆ’ 5)! = (n – 1) (n – 2) (n – 3) (n – 4) nPr = ((𝑛)!)/(𝑛 βˆ’ π‘Ÿ)! Now "nP4" /"nβˆ’1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n – 1)(n – 2)(n – 3) = 5(n – 1) (n – 2) (n – 3) (n – 4) (3𝑛(𝑛 βˆ’1)(𝑛 βˆ’2)(𝑛 βˆ’ 3))/((𝑛 βˆ’1)(𝑛 βˆ’2)(𝑛 βˆ’ 3)) = 5(n – 4) 3n = 5(n – 4) 3n = 5n – 20 20 = 5n – 3n 20 = 2n 20/2 = n 10 = n Hence, n = 10 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 Example 12 Find the value of n such that (ii) "nP4" /"nβˆ’1P4" = 5/3 , n > 4 Lets first calculate nP4 and n – 1P4 nP4 = 𝑛!/(𝑛 βˆ’ 4)! = (𝑛(𝑛 βˆ’ 1)(𝑛 βˆ’ 2)(𝑛 βˆ’ 3)(𝑛 βˆ’ 4)!)/(𝑛 βˆ’ 4)! = n(n – 1)(n – 2)(n – 3) n – 1P4 = ((𝑛 βˆ’ 1)!)/(𝑛 βˆ’ 1 βˆ’ 4)! = ((𝑛 βˆ’ 1)!)/(𝑛 βˆ’ 5)! = ((𝑛 βˆ’ 1)(𝑛 βˆ’ 2)(𝑛 βˆ’ 3)(𝑛 βˆ’ 4)(𝑛 βˆ’ 5)!)/(𝑛 βˆ’ 5)! = (n – 1) (n – 2) (n – 3) (n – 4) nPr = ((𝑛)!)/(𝑛 βˆ’ π‘Ÿ)! Now "nP4" /"nβˆ’1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n – 1)(n – 2)(n – 3) = 5(n – 1) (n – 2) (n – 3) (n – 4) (3𝑛(𝑛 βˆ’1)(𝑛 βˆ’2)(𝑛 βˆ’ 3))/((𝑛 βˆ’1)(𝑛 βˆ’2)(𝑛 βˆ’ 3)) = 5(n – 4) 3n = 5(n – 4) 3n = 5n – 20 20 = 5n – 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 12 years. He provides courses for Maths and Science at Teachoo.