Find all natural numbers $(a,b)$ such that $\displaystyle \binom{a}{b}=126$
Try: For $\displaystyle 0\leq b\leq \frac{a}{2}.$ Function $\displaystyle \binom{a}{b}$ is increasing
For $b=1$, we have $(126,1)$ and $(126,125)$
For $b=2,$ we have
$\displaystyle 126=\binom{a}{b}\geq \binom{a}{2}=\frac{a(a-1)}{2}\Rightarrow a\leq 16$
And $a!$ must be divisible by $126=2\cdot 3^2\cdot 7$
So we have $b\leq 8$
For $b=3,$ we have $\displaystyle 126=\binom{a}{b}\geq \binom{a}{3}=\frac{a(a-1)(a-2)}{6}\Rightarrow a\leq 10$
so we have $b\leq 5$
For $b=4,$ we have $\displaystyle 126=\binom{a}{b}\geq \binom{a}{4}\Rightarrow a\leq 9$
So we have $b\leq 4.5$
for $a=9$ and $b =4,$ we have $\displaystyle \binom{9}{4} = \binom{9}{5} = 126$
So all natural ordered pairs are
$(126,1),(126,125),(9,4),(9,5)$
Could someone help me to explain is what I have found is right
or any other pair exists.