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Question
If nCr–1 = 6435, nCr = 5005, nCr+1 = 3003, find rC5
Solution
nCr–1 = 6435, nCr = 5005, nCr+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
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