maths about Binomial Theorem

Let nCr=120, nC(r+1)=210, nC(r+2)=252

(n!)/(r!(n-r)!)=120
n!=120(r!(n-r)!) ..................................... (1)

(n!)/((r+1)!(n-r-1)!)=210 .......................... (2)
n!=210((r+1)!(n-r-1)!)

Put (1) into (2),
(120(r!(n-r)!))/((r+1)!(n-r-1)!)=210
(n-r)/(r+1)=210/120
n-r=(210/120)(r+1) .................................(*)

(n!)/((r+2)!(n-r-2)!)=252
n!=252((r+2)!(n-r-2)!) ............................. (3)

Put (3) into (2),
252((r+2)!(n-r-2)!)/((r+1)!(n-r-1)!)=210
(r+2)/(n-r-1)=210/252 .............................(**)

Put (*) into (**),
(r+2)/((210/120)(r+1)-1)=210/252
(-11/24)r=-11/8
r=3

by (*), n-3=(210/120)(3+1)
n=10

So, the value of n = 10