2016/102) if${n \choose r-1} = 36$, ${n \choose r} = 84$, ${n \choose r+1} = 126$ find the values of n and r

we have
$\frac{n!}{(r-1)!(n-r+1)!} = 36$, ,
$\frac{n!}{(r!(n-r)!} = 84$
$\frac{n!}{(r+1)!(n-r-1)!} = 126$
by dividing we get
$\frac{r}{n-r+1} = \frac{3}{7}$ or $7r = 3n - 3r + 3 $ or $3n- 10r + 3 = 0$
and  $\frac{r+1}{n-r} = \frac{2}{3}$ or $3(r+1) = 2(n -r)$ or $5r -2n +3=0$
multiplying 2nd by 2 and adding gto 1st we get 9-n = 0 so n = 9 and putting in (1) we get r = 3