Given
\nCr-1 = 36
\nCr+1= 126
\nC r = 84
\nC r / nCr+1 = (r+1) / (n-r)
\84/126 = (r+1) / (n-r)
\2/3 = (r+1) / (n-r)
\2(n - r) = 3(r + 1)
\2n-2r = 3r+3
\2n - 5r - 3 = 0
\5r = 2n - 3
\nCr/nCr-1 = (n-r+1) / r
\84/36 = (n-r+1) / r
\7/3 = (n-r+1) / r
\3(n-r+1) = 7r
\3n-3r+3=7r
\3n-10r+3=0
\3n-2×5r+3=0
\Substitute 5r = 2n - 3
\3n-2(2n - 3)+3=0
\3n - 4n + 6 + 3 = 0
\n = 9
\5r = 2n - 3
\Substitute n = 9
\5r = 2*9 - 3
\5r = 15
\r = 3
\Solution is r = 3
\