Example 13 class 11 ch 7 i.jpg

Example 13 class 11 ch 7 ii.jpg

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

Example 13 Find r, if 5 4Pr = 6 4Pr-1 Now it is given that 5 4Pr = 6 5Pr-1 5 ร— 4!/(4 โˆ’ ๐‘Ÿ)! = 6 ร— 5!/(5โˆ’ (๐‘Ÿ โˆ’ 1))! 5 ร— 4!/(4 โˆ’ ๐‘Ÿ)! = 6 ร— 5!/(5 โˆ’ ๐‘Ÿ + 1)! 5 ร— 4!/(4 โˆ’ ๐‘Ÿ)! = 6 ร— 5!/(6 โˆ’ ๐‘Ÿ)! (6 โˆ’ ๐‘Ÿ)!/(4 โˆ’ ๐‘Ÿ)! = (6 ร— 5!)/(5 ร— 4! ) ((6 โˆ’ ๐‘Ÿ)(5 โˆ’ ๐‘Ÿ)(4 โˆ’ ๐‘Ÿ)!)/(4 โˆ’ ๐‘Ÿ)! = (6 ร— 5!)/(5 ร— 4! ) (6 โ€“ r) (5 โ€“ r) = (6 ร— 5!)/(5 ร— 4! ) (6 โ€“ r) (5 โ€“ r) = (6 ร— 5 ร— 4!)/(5 ร— 4! ) (6 โ€“ r)(5 โ€“ r) = 6 6(5 โ€“ r) โ€“ r(5 โ€“ r) = 6 30 โ€“ 6r โ€“ 5r + r2 = 6 30 โ€“ 11r + r2 = 6 r2 โ€“ 11r + 30 = 6 r2 โ€“ 11r + 30 โ€“ 6 = 0 r2 โ€“ 11r + 24 = 0 r โ€“ 8r โ€“ 3r + 24 = 0 r(r - 8) โ€“ 3(r โ€“ 8) = 0 (r โ€“ 8) (r โ€“ 3) = 0 Hence r = 3, 8.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.