(n,k) in binomial coefficients

Calculation of integer ordered pair,s $\displaystyle (n,k)$ for which $\displaystyle \displaystyle \binom{n}{k} = 2002$

My approach:: Clearly $\displaystyle (n,k) = (2002,0)\;\;,(2002,2001)$ are the solution. Now we will find whether any ordered pair, exists OR not

$\displaystyle \displaystyle \binom{n}{k} = 2002 = 2 \times 7 \times 11 \times 13 = \frac{14 \times 13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}$

$\displaystyle \displaystyle \binom{n}{k} = \binom{14}{9} = \binom{14}{5}$

So $\displaystyle (n,k) = (14,9)\;\;,(14,5)$

But I did not understand How can I calculate any other pairs exists or not

Help me please

Thanks

Re: (n,k) in binomial coefficients

I will give you two of them: $\displaystyle \binom{2002}{1} = \binom{2002}{2001} = 2002$.

Re: (n,k) in binomial coefficients

In general, this problem is unsolved. You can look at this link: (here). It implies that the only four solutions are $\displaystyle (2002,1), (2002,2001), (14,9), (14,5)$.