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

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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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