Question

Find n, if: `((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24 : 1

Answer 1

`((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