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प्रश्न
If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then find rC2.
उत्तर
Given that nCr – 1 = 36 ......(i)
nCr = 84 ......(ii)
nCr + 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.
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