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

This video is only available for Teachoo black users


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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.