Question

If ⁿP_r = 1814400 and ⁿC_r = 45, find ⁿ⁺⁴C_r+3 

Answer 1

ⁿP_r = 1814400, ⁿC_r = 45

∴ `(""^"n""P"_"r")/(""^"n""C"_"r") = 1814400/45`

∴ `("n"!)/(("n" - "r")!) xx ("r"!("n" - "r")!)/("n"!) = 1814400/45`

∴ r! = 40320

∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

∴ r! = 8!

∴ r = 8

Also, ⁿC_r = 45

∴ ⁿC₈ = 45

∴ `("n"!)/(8!("n" - 8)!)` = 45

∴ n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5)(n – 6) (n – 7) = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3

Comparing on both sides, we get

n = 10

∴ ⁿ⁺⁴C_r+3 = ¹⁴C₁₁

= `(14!)/(11!(14 - 11)!)`

= `(14!)/(11!3!)`

= `(14 xx 13 xx 12 xx 11!)/(11! xx 3 xx 2 xx 1)`

= 364