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  1. Chapter 7 Class 11 Permutations and Combinations
  2. Serial order wise

Transcript

Ex 7.3, 7 Find r if 5Pr = 2 6Pr โ€“ 1 Calculating 5Pr & 6Pr โ€“ 1 Given, 5Pr = 26Pr-1 5Pr = 5!/(5 โˆ’ ๐‘Ÿ)! 6Pr โ€“ 1 = 6!/(6 โˆ’(๐‘Ÿโˆ’1))! = 6!/(6 โˆ’ ๐‘Ÿ + 1)! = 6!/(7 โˆ’ ๐‘Ÿ)! nPr = ๐‘›!/(๐‘› โˆ’ ๐‘Ÿ)! 5!/(5 โˆ’ ๐‘Ÿ)! = 2 ร— 6!/(7 โˆ’ ๐‘Ÿ)! ((7 โˆ’ ๐‘Ÿ)! )/(5 โˆ’ ๐‘Ÿ)! = 2 ร— 6!/5! ((7 โˆ’ ๐‘Ÿ)(6 โˆ’ ๐‘Ÿ)(5 โˆ’ ๐‘Ÿ)! )/(5 โˆ’ ๐‘Ÿ)! = 2 ร— 6!/5! (7 โ€“ r)(6 โ€“ r) = 2 ร— 6!/5! (7 โ€“ r)(6 โ€“ r) = 2 ร— (6 ร— 5!)/5! (7 โ€“ r)(6 โ€“ r) = 2 ร— 6 (7 โ€“ r)(6 โ€“ r) = 12 7(6 โ€“ r) โ€“ r(6 โ€“ r) = 12 42 โ€“ 7r โ€“ 6r + r2 = 12 r2 โ€“ 13r + 42 โ€“ 12 = 0 r2 โ€“ 13r + 30 = 0 r2 โ€“ 3r โ€“ 10r + 30 = 0 r (r โ€“ 3) โ€“ 10 (r โ€“ 3) = 0 (r โ€“ 3) (r โ€“ 10) = 0 Hence, r = 3, 10 But, r < n So, r < 5 and r < 6 โˆด r = 10 is not possible So, r = 3 is the answer Ex 7.3, 7 Find r if (ii) 5Pr = 6Pr โ€“ 1 Calculating 5Pr & 6Pr โ€“ 1 Given 5Pr = 6Pr โ€“ 1 5Pr = 5!/(5 โˆ’ ๐‘Ÿ)! 6Pr โ€“ 1 = 6!/(6 โˆ’ (๐‘Ÿ โˆ’ 1))! = 6!/(6 โˆ’ ๐‘Ÿ + 1)! = 6!/(7 โˆ’ ๐‘Ÿ)! nPr = ((๐‘›)!)/(๐‘› โˆ’ ๐‘Ÿ)! 5!/(5 โˆ’ ๐‘Ÿ)! = 6!/(7 โˆ’ ๐‘Ÿ)! ((7 โˆ’ ๐‘Ÿ)! )/(5 โˆ’ ๐‘Ÿ)! = 6!/5! ((7 โˆ’ ๐‘Ÿ)(6 โˆ’ ๐‘Ÿ)(5 โˆ’ ๐‘Ÿ)! )/(5 โˆ’ ๐‘Ÿ)! = 6!/5! (7 โ€“ r)(6 โ€“ r) = 6!/5! (7 โ€“ r)(6 โ€“ r) = (6 ร— 5!)/5! (7 โ€“ r)(6 โ€“ r) = 6 7(6 โ€“ r) โ€“ r(6 โ€“ r) = 6 42 โ€“ 7r โ€“ 6r + r2 = 6 r2 โ€“ 13r + 42 โ€“ 6 = 0 r2 โ€“ 13r + 36 = 0 r2 โ€“ 9r โ€“ 4r + 36 = 0 r (r โ€“ 9) โ€“ 4 (r โ€“ 9) = 0 (r โ€“ 4) (r โ€“ 9) = 0 Hence r = 4, 9 But, r < n So, r < 5 and r < 6 โˆด r = 9 is not possible So, r = 4 is the answer

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.