Ex 6.4, 2 (ii) - Chapter 6 Class 11 Permutations and Combinations

Last updated at Dec. 16, 2024 by

Slide5.JPG

Slide6.JPG

Go Ad-free

Transcript

Ex 6.4, 2 Determine n if (ii) 2nC3 : nC3 = 11 : 1 Let first calculate 2nC3 and nC3 separately 2nC3 = 2 !/3!(2 3)! = ((2 )(2 1)(2 2)(2 3)!)/((3 2 1)(2 3)!) = 2 (2 1)(2 2)/6 Given 2nC3 : nc3 = 11 : 1 2 (2 1)(2 2)/6 : ( 1)( 2)/6 = 11 : 1 2 (2 1)(2 2)/6 : ( 1)( 2)/6 = 11 : 1 ((2 (2 1)(2 2))/6)/( ( 1)( 2)/6) = 11 2n(2n 1)(2n 2)/6 6/( ( 1)( 2)) = 11 2n(2n 1)(2n 2)/( ( 1)( 2)) 6/6 = 11 (2(2n 1). 2(n 1))/(( 1)( 2)) = 11 4(2n 1)(n 1)/(( 1)( 2)) = 11 4(2n 1)/(( 2)) = 11 4(2n 1) = 11(n 2) 8n 4 = 11n 22 4 + 22 = 11n 8n 18 = 3n 18/3 = n 6 = n Thus, n = 6

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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo