Question

If ⁿC_{r – 1} = 36, ⁿC_r = 84 and ⁿC_{r + 1} = 126, then find ^rC₂.

Answer 1

Given that ⁿC_{r – 1} = 36  ......(i)

ⁿC_r = 84  ......(ii)

ⁿC_{r + 1} = 126   ......(iii)

Dividing equation (i) by equation (ii) we get

`(""^n"c"_(r - 1))/(""^n"C"_1) = 36/84`

⇒ `((n!)/((r - 1)!(n - r + 1)!))/((n!)/(r!(n - r)!)) = 3/7`  .......`[because ""^n"C"_r = (n!)/(r!(n - r)!)]`

⇒ `(n!)/((r - 1)!(n - r + 1)!) xx (r!(n - r)!)/(n1) = 3/7`

⇒ `(r*(r - 1)!(n - r)!)/((r - 1)!(n - r + 1)(n - r)!) = 3/7`

⇒ `r/(n - r + 1) = 3/7`

⇒ 3n – 3r + 3 = 7r

⇒ 3n – 10r = – 3   ......(iv)

Now dividing equation (ii) by equation (iii), we get

`(""^n"C"_r)/(""^n"C"_(r + 1)) = 84/126`

⇒ `((n1)/(r!(n - r)!))/((n!)/((r + 1)!(n - r - 1)!)) = 2/3`

⇒ `(n!)/(r!(n - r)!) xx ((r + 1)! (n - r - 1)1)/(n!) = 2/3`

⇒ `((r + 1) * r!(n - r - 1)!)/(r!(n - r)(n - r - 1)!) = 2/3`

⇒ `(r + 1)/(n - r) = 2/3`

⇒ 2n – 2r = 3r + 3

⇒ 2n – 5r = 3  ....(v)

Solving equation (iv) and (v) we have

     3n – 10r = – 3
       2n – 5r = 3
     3n – 10r = – 3
     4n – 10r =    6
(–)    (+)      (–)      
  – n = – 9 ⇒ n = 9

∴ 2 × 9 – 5r = 3

⇒ 18 – 5r = 3

⇒ r = `15/5` = 3

So, ^rC₂ = ³C2

= `(3!)/(2!(3 - 2)!)` = 3

Hence, the value of ^rC₂ = 3.