Question

Find n and r if ⁿP_r = 720 and ⁿC_n–r = 120

Answer 1

ⁿP_r = 720 

∴ `("n"!)/(("n" - "r")!)` = 720     ...(i)

Also, ⁿC_n–r = 120

∴ `("n"!)/(("n" - "r")!("n" - "n" + "r")!)` = 120

∴ `("n"!)/("r"!("n" - "r")!)` = 120   ...(ii)

Dividing (i) by (ii), we get

`(("n"!)/(("n" - "r")!))/(("n"!)/("r"!("n" - "r")!)) = 720/120`

∴ r! = 6

∴ r = 3

Substituting r = 3 in (i), we get

`("n"!)/(("n" - 3)!)` = 720

∴ `("n"("n" - 1)("n" - 2)("n" - 3)!)/(("n" - 3)!)` = 720

∴ n(n – 1) (n – 2) = 10 × 9 × 8

∴ n = 10