Question

If `""^(("n" + 1))"C"_8 : ""^(("n" - 3))"P"_4` = 57 : 16, find the value of n

Answer 1

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`

ⁿP_r = `("n"!)/(("n" - "r")!)`, ⁿC_r = `("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