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Question
If `""^(("n" + 1))"C"_8 : ""^(("n" - 3))"P"_4` = 57 : 16, find the value of n
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Solution
Given `""^(("n" + 1))"C"_8 : ""^(("n" - 3))"P"_4` = 57 : 16
`(""^(("n" + 1))"C"_8)/(""^(("n" - 3))"P"_4) = 57/16`
`16""^(("n" + 1))""_8 = 57""^(("n" - 3))"P"_4`
nPr = `("n"!)/(("n" - "r")!)`, nCr = `("n"!)/("r"!("n" - "r")!)`
`16""^(("n" + 1))"C"_8 = 57""^(("n" - 3))"P"_4`
`16 xx (("n" + 1)!)/(8!("n" + 1 - 8)!) = 57 xx (("n" - 3)!)/(("n" - 3- 4)!)`
`16 xx (("n" + 1)!)/(8!("n"- 7)!) = 57 xx (("n" - 3)!)/(("n" - 7)!)`
`16 xx (("n" + 1)!)/(8!) = 57 xx ("n" - 3)!`
16 × (n + 1)(n)(n – 1)(n – 2)(n – 3)! = 57 × 8! (n – 3)!
(n + 1)n(n – 1)(n – 2) = `(57 xx 8!)/16`
(n + 1)n(n – 1)(n – 2) = `(57 xx 8 xx 7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1)/16`
(n + 1)n(n – 1)(n – 2) = 57 × 7 × 6 × 5 × 4 × 3
= 3 × 19 × 7 × 6 × 5 × 4 × 3
= (7 × 3) × (5 × 4)19 × (6 × 3)
= 21 × 20 × 19 × 18
= (20 + 1) × (20) × (20 – 1) × (20 – 2)
∴ n = 20
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