Example 13 - Chapter 6 Class 11 Permutations and Combinations
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 13 Find r, if 5 4Pr = 6 4Pr-1 Now it is given that 5 4Pr = 6 5Pr-1 5 ร 4!/(4 โ ๐)! = 6 ร 5!/(5โ (๐ โ 1))! 5 ร 4!/(4 โ ๐)! = 6 ร 5!/(5 โ ๐ + 1)! 5 ร 4!/(4 โ ๐)! = 6 ร 5!/(6 โ ๐)! (6 โ ๐)!/(4 โ ๐)! = (6 ร 5!)/(5 ร 4! ) ((6 โ ๐)(5 โ ๐)(4 โ ๐)!)/(4 โ ๐)! = (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. But, r < n So, r < 4 and r < 5 โด r = 8 is not possible So, r = 3 is the answer
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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