Question

Find n and r if ⁿC_r–1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42

Answer 1

ⁿC_r–1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42

∴ ⁿC_r–1 : ⁿC_r = 20 : 35 and

ⁿC_r : ⁿC_r+1 = 35 : 42

Take, ⁿC_r–1 : ⁿC_r = 20 : 35

∴ `("n"!)/(("r" - 1)! ("n" - "r" + 1)!) xx ("r"!("n" - "r")!)/("n"!) = 20/35`

∴ `1/(("r" - 1)!("n" - "r" + 1)("n" - "r")!) xx "r"("r" - 1)!("n" - "r")! = 20/35`

∴ `"r"/("n" - "r" + 1) = 20/35`

∴ 35r = 20n – 20r + 20

∴ 55r – 20n = 20

∴ 11r – 4n = 4    ...(1)

Take ⁿC_r : ⁿC_r+1 = 35 : 42

∴ `("n"!)/("r"!("n" - "r")!) xx (("r" + 1)! ("n" - "r" - 1)!)/("n"!) = 35/42`

∴ `(("r" + 1) xx "r"! xx ("n" - "r" - 1)!)/("r"!("n" - "r")("n" - "r" - 1)!) = 35/42`

∴ `("r" + 1)/("n" - "r") = 5/6`

∴ 6r + 6 = 5n – 5r

∴ 11r – 5n = – 6   ...(2)

Subtracting equation (2) from equation (1), we get,

  11r – 4n =    4
  11r – 5n = – 6
 –     +         +     
∴           n =   10

Putting n = 10 in the equation (1), we get,

11r – 4(10) = 4

∴ 11r – 40 = 4

∴ 11r = 44

∴ r = 4

Hence, n = 10 and r = 4.