Ex 7.4, 2 - Chapter 7 Class 11 Permutations and Combinations - Part 3

Ex 7.4, 2 - Chapter 7 Class 11 Permutations and Combinations - Part 4

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.