Find $(n,r)$ such that $C(n,r) =1984$

Question

I explored this problem a bit, trying various values of $r$

$r=0 \implies C(n,0) =1984$ which is impossible

$r=1 \implies C(n,1)=1984$ which yields our first solution (and probably the only one) $(n,r)=(1984,1)$

$r=2 \implies C(n,2)=1984 \iff n(n-1)=2\cdot 1984=3968$

Which is asking if there exists $2$ consecutive numbers that multiply out to $3968$

And this seems to continue for any $r>1$

So we can maybe reform the question into the language of number theory, if there exists $r$ consecutive numbers $n, n-1, ... , n-r+1$ such that $n(n-1)...(n-r+1)=r!\cdot1984$

There's a probably some other route that has to do with showing after a certain value of $r$, there are no other solutions too.

Answer 1

Comment

${32 \choose 2}=\frac{31\times 32}2$

${32 \choose 3}=5\times 31\times 32=2\times 64\times 31+2\times\frac{31\times 32}2$

$\Rightarrow 2\times 1984+2{32 \choose 2}={32 \choose 3}\rightarrow 1984=\frac12 [{32 \choose 3}+{32 \choose 2}]-\frac32{32 \choose 2}$

Using formula ${n+1 \choose r}={n \choose r}+{n \choose r-1}$, we obtain:

$1984=\frac12{33 \choose 3}-\frac32{32 \choose 2}$

So it seems that the is no integer solution except n=1984 and r=1.