Find all values of $n, m \in \mathbb{Z^+}$ where ${n \choose m}=1984$

Question

Find all $n, m \in \mathbb{Z^+}$ where ${n \choose m}=1984$

I tried so hard on this question in the past couple of days I couldn't achieve anything, which forced me to resort to the guess and check method, but even that was hard. I tried to convert it to $\frac{n!}{m!(n-m)!}=1984$ but couldn't reach anything. I also attempted to find $1984=x!$ but there was no value for $x$. I was wondering if I could receive some help on this question if possible. Thank you anyways.

Answer 1

Since $31\mid 1984$, $n\geq 31$ must hold. If $m\geq 3$ or $n-m\geq 3$, then $$\binom{n}{m}\geq \binom{n}{3} \geq \binom{31}{3}=4495>1984\,.$$ Therefore, $m\leq 2$ or $n-m\leq 2$. Thus, you are left with not so many choices now.

All solutions are $(n,m)=(1984,1)$ and $(n,m)=(1984,1983)$.