Question

If ⁿC_r–1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003, find ^rC₅

Answer 1

ⁿC_r–1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003

∴ `(""^"n""C"_("r"-1))/(""^"n""C"_"r") = 6435/5005 = 9/7`

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

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

∴ `"r"/("n" - "r" + 1) = 9/7`

∴ 7r = 9n – 9r + 9

∴ 16r – 9n = 9     ...(1)

Also, `(""^"n""C"_"r")/(""^"n""C"_("r" + 1)) = 5005/3003 = 5/3`

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

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

∴ `("r" + 1)/("n" - "r") = 5/3`

∴ 3r + 3 = 5n − 5r

∴ 8r – 5n = – 3   ...(2)

Multiplying equation (1) by 5 and equation (2) by 9, we get,

80r – 45n = 45

72r – 45n = – 27

On subtracting, we get,

8r = 72

∴ r = 9

∴ ^rC₅ = ⁹C₅ = `(9!)/(5!4!)`

= `(9 xx 8 xx 7 xx 6 xx 5!)/(5! xx 4 xx 3 xx 2 xx1)`

= 126