(n,k) in binomial coefficients

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• October 14th 2013, 07:26 PM
jacks
(n,k) in binomial coefficients
Calculation of integer ordered pair,s $(n,k)$ for which $\displaystyle \binom{n}{k} = 2002$

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

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

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

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

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

Help me please

Thanks
• October 14th 2013, 07:32 PM
SlipEternal
Re: (n,k) in binomial coefficients
I will give you two of them: $\binom{2002}{1} = \binom{2002}{2001} = 2002$.
• October 14th 2013, 08:08 PM
SlipEternal
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 $(2002,1), (2002,2001), (14,9), (14,5)$.