If n^cr-1=36 and n^cr=84 and n^cr+1=126

Question

then r =_______

Answer 1

Given

ⁿC_r-1 = 36

ⁿC_r+1= 126

ⁿC _r  = 84

ⁿC _r / ⁿC_r+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

ⁿC_r/ⁿC_r-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