Question

If `""^"n""C"_("r" - 1)` = 6435, `""^"n""C"_"r"` = 5005, `""^"n""C"_("r" + 1)` = 3003, find `""^"r""C"_5`.

Answer 1

Given: `""^"n""C"_("r" - 1) = 6435, ""^"n""C"_"r" = 5005, ""^"n""C"_("r" + 1)` = 3003

∴  `(""^"n""C"_("r" - 1))/(""^"n""C"_"r") = 6435/5005`

∴ `(("n"!)/(["n" - ("r" - 1)]!("r" - 1)!))/(("n"!)/(("n" - "r")!"r"!)) =9/7`

∴ `("n"!)/(("n" - "r" + 1)!("r" - 1)!)xx(("n" - "r")!"r"!)/("n"!) =9/7`

∴ `("n"!)/(("n" - "r" + 1)("n"-"r")!("r" - 1)!)xx(("n" - "r")!"r"("r"-1))/("n"!) =9/7`

∴ `"r"/("n" - "r" + 1) = 9/7`

∴ 7r = 9n – 9r + 9
∴ 16r – 9n = 9       ...(I)
Also,
∴ `(""^"n""C"_"r")/(""^"n""C"_("r" + 1)) = 5005/3003`

∴ `(("n"!)/(("n" - "r")!r!))/(("n"!)/(["n" - ("r" + 1)]!("r" + 1)!)) = (5 xx 1001)/(3 xx 1001)`

∴ `(("n" - "r" - 1)!("r" + 1)!)/(("n" - "r")!"r"!) = 5/3`

∴ `(("n" - "r" - 1)!("r" + 1)"r"!)/(("n" - "r")("n" - "r" - 1)!"r"!) = 5/3`

∴ `(("r" + 1))/(("n" - "r")) = 5/3`

∴ 3(r + 1) = 5(n – r)
∴ 3r + 3 = 5n  –  5r
∴ 8r – 5n = – 3      ...(II)
Multiplying equations (II), by 2 and
Subtracting equation (I) from (III)

  16r – 10n = – 6              ...(III)
  16r – 9n = 9                   ...(I)
        +         –       
      – n =  – 15

∴ n = 15
Substituting n = 15 in equation (I)
16r – 9(15) = 9
∴ 16r – 135 = 9
∴ 16r = 144

∴ r = `144/16`
∴ r = 9
∴ `""^"r""C"_5 = ""^9"C"_5 = (9!)/(4!5!) = (9 xx 8 xx 7 xx 6 xx 5!)/(4 xx 3 xx 2 xx 1 xx 5!)`

∴ `""^"r""C"_5 = ""^9"C"_5` = 126