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Example 13 class 11 ch 7 i.jpg

Example 13 - Chapter 7 Class 11 Permutations and Combinations - Part 2
Example 13 - Chapter 7 Class 11 Permutations and Combinations - Part 3


Transcript

Example 13 Find r, if 5 4Pr = 6 4Pr-1 Now it is given that 5 4Pr = 6 5Pr-1 5 Γ— 4!/(4 βˆ’ π‘Ÿ)! = 6 Γ— 5!/(5βˆ’ (π‘Ÿ βˆ’ 1))! 5 Γ— 4!/(4 βˆ’ π‘Ÿ)! = 6 Γ— 5!/(5 βˆ’ π‘Ÿ + 1)! 5 Γ— 4!/(4 βˆ’ π‘Ÿ)! = 6 Γ— 5!/(6 βˆ’ π‘Ÿ)! (6 βˆ’ π‘Ÿ)!/(4 βˆ’ π‘Ÿ)! = (6 Γ— 5!)/(5 Γ— 4! ) ((6 βˆ’ π‘Ÿ)(5 βˆ’ π‘Ÿ)(4 βˆ’ π‘Ÿ)!)/(4 βˆ’ π‘Ÿ)! = (6 Γ— 5!)/(5 Γ— 4! ) (6 – r) (5 – r) = (6 Γ— 5!)/(5 Γ— 4! ) (6 – r) (5 – r) = (6 Γ— 5 Γ— 4!)/(5 Γ— 4! ) (6 – r)(5 – r) = 6 6(5 – r) – r(5 – r) = 6 30 – 6r – 5r + r2 = 6 30 – 11r + r2 = 6 r2 – 11r + 30 = 6 r2 – 11r + 30 – 6 = 0 r2 – 11r + 24 = 0 r – 8r – 3r + 24 = 0 r(r - 8) – 3(r – 8) = 0 (r – 8) (r – 3) = 0 Hence r = 3, 8. But, r < n So, r < 4 and r < 5 ∴ r = 8 is not possible So, r = 3 is the answer

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.