If n+2C8 : n-2P4 = 57:16, then find n - Maths - Permutations and Combinations

Question

If n+2C8 : n-2P4 =
57:16, then find n

Shivam Mishra · Accepted Answer

Solution :

Formula: NCR=N!R! N-R! and NPR=N!N-R!       n+2C8n-2P4=5716⇒n+2!8! n+2-8!n-2!n-2-4!=5716⇒n+2!8! n-6!n-2! n-6!=5716⇒n+2!8! n-2!1=5716⇒n+2!n-2!=5716×8!⇒n+2n+1nn-1n-2!n-2!=5716×8×7×6×5×4×3×2×1⇒n+2n+1nn-1=57×7×6×5×4×3As L
H S
 is product of four consecutive natural numbers, hencetry to write R
H S
 as product of four natural numbers
⇒n+2n+1nn-1=19×3×7×6×5×4×3⇒n+2n+1nn-1=19×3×7×4×5×6×3⇒n+2n+1nn-1=19×21×20×18Rearranging we get:⇒n+2n+1nn-1=21×20×19×18⇒n+2=21⇒n=21-2⇒n=19 Answer

If n+2C8 : n-2P4 = 57:16, then find n.

If ⁿ⁺²C₈ : ^n-2P₄ = 57:16, then find n.