If `""^(n)C_(r -1) = 36, ""^(n)C_(r) = 84 " and " ""^(n)C_(r +1) = 126`, then find the value of `""^(r)C_(2)`.

Explanation

`(.^(n)C_(r-1))/(.^(n)C_(r)) = (36)/(84)`
`rArr (n!)/((r-1)!(n-r+1)!).(r!(n-r)!)/(n!)=(3)/(7)`
`rArr (r(r-1)!(n-r)!)/((r-1)!(n-r+1).(n-r)!) = (3)/(7)`
`rArr (r)/(n-r+1) = (3)/(7)`
`rArr 3n - 3r +3 = 7r`
`rArr n = (10r-3)/(3)` ...(1)
and `(.^(n)C_(r))/(.^(n)C_(r+1)) = (84)/(126)`
`rArr (n!)/(r!(n-r)!)((r+1)!(n-r-1)!)/(n!) = (2)/(3)`
`rArr ((r+1).r!(n-r-1)!)/(r!(n-r)(n-r-1)!) = (2)/(3)`
`rArr (r+1)/(n-r) = (2)/(3)`
`rArr 2n - 2r = 3r +3`
`rArr n = (5r+3)/(2)` ...(2)
From eqs. (1) and (2), we get
`(5r +3)/(2) = (10r-3)/(3)`
`rArr 20r - 6 = 15r +9`
`rArr r = 3`
Now `.^(r)C_(2) = .^(3)C_(2)`
`= (3!)/(2!.1!) = 3`.