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Question
Find n, if: `((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24 : 1
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Solution
`((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24 : 1
∴ `((2"n")!)/(7!(2"n" - 7)!) xx (4!("n" - 4)!)/("n"!)` = 24
∴ `((2"n")(2"n" - 1)(2"n" - 2)(2"n" - 3)(2"n" - 4)(2"n" - 5)(2"n" - 6)(2"n" - 7)!)/(7 xx 6 xx 5 xx 4!(2"n" - 7)!) xx (4!("n" - 4)!)/("n"("n" - 1)("n" - 2)("n" - 3)("n" - 4)!)` = 24
∴ `((2"n")(2"n" - 1)(2"n" - 2)(2"n" - 3)(2"n" - 4)(2"n" - 5)(2"n" - 6))/(7 xx 6 xx 5) xx 1/("n"("n" - 1)("n" - 2)("n" - 3)` = 24
∴ `((2"n")(2"n" - 1)2("n" - 1)(2"n" - 3)2("n" - 2)(2"n" - 5)2("n" - 3))/(7 xx 6 xx 5) xx 1/("n"("n" - 1)("n" - 2)("n" - 3)` = 24
∴ `(16(2"n" - 1)(2"n" - 3)(2"n" - 5))/(7 xx 6 xx 5)` = 24
∴ (2n – 1) (2n – 3) (2n – 5) = `(24 xx 7 xx 6 xx 5)/16`
∴ (2n – 1) (2n – 3) (2n – 5) = 9 × 7 × 5
Comparing on both sides, we get
∴ 2n – 1 = 9
∴ n = 5
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